ID: 674 Type: Subjective Source: NCERT Class 11

Question 1: Find the equation of the circle with centre $(0,2)$ and radius $2$.

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🤖 AI Generated Solution:

The standard equation of a circle with centre $(h, k)$ and radius $r$ is given by $(x - h)^2 + (y - k)^2 = r^2$. Given centre $(h, k) = (0, 2)$ and radius $r = 2$. Substitute these values into the standard equation: $(x - 0)^2 + (y - 2)^2 = 2^2$ $x^2 + (y - 2)^2 = 4$ Thus, the equation of the circle is $x^2 + (y - 2)^2 = 4$.

ID: 675 Type: Subjective Source: NCERT Class 11

Question 2: Find the equation of the circle with centre $(-2,3)$ and radius $4$.

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🤖 AI Generated Solution:

The standard equation of a circle with centre $(h,k)$ and radius $r$ is given by $(x-h)^2 + (y-k)^2 = r^2$. Given centre $(h,k) = (-2,3)$ and radius $r = 4$. Substitute these values into the standard equation: $(x - (-2))^2 + (y - 3)^2 = 4^2$ $(x + 2)^2 + (y - 3)^2 = 16$ Thus, the equation of the circle is $(x + 2)^2 + (y - 3)^2 = 16$.

ID: 676 Type: Subjective Source: NCERT Class 11

Question 3: Find the equation of the circle with centre $(\frac{1}{2}, \frac{1}{4})$ and radius $\frac{1}{12}$.

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🤖 AI Generated Solution:

The standard equation of a circle with centre $(h, k)$ and radius $r$ is given by: $$(x - h)^2 + (y - k)^2 = r^2$$ Given centre $(h, k) = (\frac{1}{2}, \frac{1}{4})$ and radius $r = \frac{1}{12}$. Substitute these values into the standard equation: $$(x - \frac{1}{2})^2 + (y - \frac{1}{4})^2 = (\frac{1}{12})^2$$ $$(x^2 - x + \frac{1}{4}) + (y^2 - \frac{1}{2}y + \frac{1}{16}) = \frac{1}{144}$$ $x^2 + y^2 - x - \frac{1}{2}y + \frac{1}{4} + \frac{1}{16} - \frac{1}{144} = 0$ To combine constant terms, find a common denominator, which is $144$: $x^2 + y^2 - x - \frac{1}{2}y + \frac{36}{144} + \frac{9}{144} - \frac{1}{144} = 0$ $x^2 + y^2 - x - \frac{1}{2}y + \frac{44}{144} = 0$ Simplify the fraction $\frac{44}{144}$ to $\frac{11}{36}$: $x^2 + y^2 - x - \frac{1}{2}y + \frac{11}{36} = 0$ Multiplying the entire equation by $36$ to clear denominators: $$36x^2 + 36y^2 - 36x - 18y + 11 = 0$$

ID: 677 Type: Subjective Source: NCERT Class 11

Question 4: Find the equation of the circle with centre $(1,1)$ and radius $\sqrt{2}$.

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🤖 AI Generated Solution:

The standard equation of a circle with centre $(h,k)$ and radius $r$ is given by: $$(x-h)^2 + (y-k)^2 = r^2$$ Given the centre $(h,k) = (1,1)$ and radius $r = \sqrt{2}$. Substitute these values into the standard equation: $$(x-1)^2 + (y-1)^2 = (\sqrt{2})^2$$ Simplifying the right-hand side gives: $$(x-1)^2 + (y-1)^2 = 2$$ Expanding this equation further, we get: $$x^2 - 2x + 1 + y^2 - 2y + 1 = 2$$ $$x^2 + y^2 - 2x - 2y + 2 = 2$$ $$x^2 + y^2 - 2x - 2y = 0$$ The equation of the circle is either $(x-1)^2 + (y-1)^2 = 2$ or $x^2 + y^2 - 2x - 2y = 0$.

ID: 678 Type: Subjective Source: NCERT Class 11

Question 5: Find the equation of the circle with centre $(-a, -b)$ and radius $\sqrt{a^{2}-b^{2}}$.

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🤖 AI Generated Solution:

The standard equation of a circle with centre $(h, k)$ and radius $r$ is given by $(x-h)^2 + (y-k)^2 = r^2$. Given: Centre $(h, k) = (-a, -b)$ Radius $r = \sqrt{a^2-b^2}$ 1. Substitute the given centre and radius into the standard equation: $(x - (-a))^2 + (y - (-b))^2 = (\sqrt{a^2-b^2})^2$ 2. Simplify the terms: $(x+a)^2 + (y+b)^2 = a^2-b^2$ 3. Expand the squared terms: $(x^2 + 2ax + a^2) + (y^2 + 2by + b^2) = a^2-b^2$ 4. Rearrange and combine like terms to get the general form of the equation: $x^2 + y^2 + 2ax + 2by + a^2 + b^2 - (a^2-b^2) = 0$ $x^2 + y^2 + 2ax + 2by + a^2 + b^2 - a^2 + b^2 = 0$ $x^2 + y^2 + 2ax + 2by + 2b^2 = 0$ The equation of the circle is $\boxed{x^2 + y^2 + 2ax + 2by + 2b^2 = 0}$.

ID: 679 Type: Subjective Source: NCERT Class 11

Question 6: Find the centre and radius of the circle $(x+5)^{2}+(y-3)^{2}=36$.

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🤖 AI Generated Solution:

The standard equation of a circle is given by $(x-h)^2 + (y-k)^2 = r^2$, where $(h,k)$ is the centre and $r$ is the radius. Comparing the given equation $(x+5)^{2}+(y-3)^{2}=36$ with the standard form: We can rewrite $(x+5)^2$ as $(x-(-5))^2$. Thus, $h = -5$ and $k = 3$. The value for $r^2$ is $36$, so $r = \sqrt{36} = 6$ (since radius must be positive). Therefore, the centre of the circle is $(-5, 3)$ and the radius is $6$.

ID: 680 Type: Subjective Source: NCERT Class 11

Question 7: Find the centre and radius of the circle $x^{2}+y^{2}-4x-8y-45=0$.

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🤖 AI Generated Solution:

To find the centre and radius of the circle, we convert the given equation into the standard form $(x-h)^2 + (y-k)^2 = r^2$. The given equation is $x^{2}+y^{2}-4x-8y-45=0$. 1. Group $x$ terms and $y$ terms, and move the constant to the right side: $x^{2}-4x + y^{2}-8y = 45$ 2. Complete the square for the $x$ terms and $y$ terms. Add $(\frac{-4}{2})^2 = 4$ for $x$ and $(\frac{-8}{2})^2 = 16$ for $y$ to both sides of the equation: $x^{2}-4x+4 + y^{2}-8y+16 = 45+4+16$ 3. Rewrite the perfect squares: $(x-2)^2 + (y-4)^2 = 65$ 4. Comparing this to the standard form $(x-h)^2 + (y-k)^2 = r^2$, we identify the centre $(h,k)$ and the radius $r$: Centre: $(2, 4)$ Radius: $r^2 = 65 \Rightarrow r = \sqrt{65}$ The centre of the circle is $(2, 4)$ and the radius is $\sqrt{65}$.

ID: 681 Type: Subjective Source: NCERT Class 11

Question 8: Find the centre and radius of the circle $x^{2}+y^{2}-8x+10y-12=0$.

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🤖 AI Generated Solution:

To find the centre and radius of the circle, we compare the given equation with the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$. The given equation is $x^{2}+y^{2}-8x+10y-12=0$. 1. **Identify coefficients**: Comparing, we have $2g = -8$, so $g = -4$. We have $2f = 10$, so $f = 5$. We have $c = -12$. 2. **Calculate the Centre**: The centre of the circle is given by $(-g, -f)$. Centre $= (-(-4), -(5)) = (4, -5)$. 3. **Calculate the Radius**: The radius of the circle is given by $r = \sqrt{g^2 + f^2 - c}$. $r = \sqrt{(-4)^2 + (5)^2 - (-12)}$ $r = \sqrt{16 + 25 + 12}$ $r = \sqrt{53}$ Thus, the centre of the circle is $(4, -5)$ and the radius is $\sqrt{53}$.

ID: 682 Type: Subjective Source: NCERT Class 11

Question 9: Find the centre and radius of the circle $2x^{2}+2y^{2}-x=0$.

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🤖 AI Generated Solution:

The equation of the circle is $2x^{2}+2y^{2}-x=0$. 1. **Standardize the equation**: Divide the entire equation by 2 to make the coefficients of $x^2$ and $y^2$ equal to 1. $x^{2}+y^{2}-\frac{1}{2}x=0$ 2. **Compare with general form**: The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. Comparing, we get: $2g = -\frac{1}{2} \implies g = -\frac{1}{4}$ $2f = 0 \implies f = 0$ $c = 0$ 3. **Find the centre**: The centre of the circle is given by $(-g, -f)$. Centre $C = \left(-\left(-\frac{1}{4}\right), -0\right) = \left(\frac{1}{4}, 0\right)$ 4. **Find the radius**: The radius $r$ is given by the formula $r = \sqrt{g^2 + f^2 - c}$. $r = \sqrt{\left(-\frac{1}{4}\right)^2 + (0)^2 - 0} = \sqrt{\frac{1}{16} + 0 - 0} = \sqrt{\frac{1}{16}} = \frac{1}{4}$ The centre of the circle is $\left(\frac{1}{4}, 0\right)$ and the radius is $\frac{1}{4}$.

ID: 683 Type: Subjective Source: NCERT Class 11

Question 10: Find the equation of the circle passing through the points $(4,1)$ and $(6,5)$ and whose centre is on the line $4x+y=16$.

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🤖 AI Generated Solution:

Let the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$, where $(h,k)$ is the center and $r$ is the radius. 1. Since the center $(h,k)$ lies on the line $4x+y=16$, we have: $4h+k=16 \quad \text{(1)}$ 2. The circle passes through $(4,1)$ and $(6,5)$. The distance from the center to these points must be equal to the radius $r$. $(4-h)^2 + (1-k)^2 = r^2$ $(6-h)^2 + (5-k)^2 = r^2$ 3. Equating the expressions for $r^2$: $(4-h)^2 + (1-k)^2 = (6-h)^2 + (5-k)^2$ $16 - 8h + h^2 + 1 - 2k + k^2 = 36 - 12h + h^2 + 25 - 10k + k^2$ $17 - 8h - 2k = 61 - 12h - 10k$ $4h + 8k = 44 \implies h+2k=11 \quad \text{(2)}$ 4. Solve the system of equations (1) and (2) for $h$ and $k$: From (1), $k=16-4h$. Substitute into (2): $h+2(16-4h)=11$ $h+32-8h=11$ $-7h = -21 \implies h=3$ Substitute $h=3$ into $k=16-4h$: $k=16-4(3) = 16-12 = 4$ The center of the circle is $(3,4)$. 5. Calculate $r^2$ using the center $(3,4)$ and point $(4,1)$: $r^2 = (4-3)^2 + (1-4)^2 = 1^2 + (-3)^2 = 1+9 = 10$. The equation of the circle is $(x-3)^2 + (y-4)^2 = 10$.

ID: 684 Type: Subjective Source: NCERT Class 11

Question 11: Find the equation of the circle passing through the points $(2,3)$ and $(-1,1)$ and whose centre is on the line $x-3y-11=0$.

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🤖 AI Generated Solution:

Let the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$. 1. Since the circle passes through $(2,3)$ and $(-1,1)$: $(2-h)^2 + (3-k)^2 = r^2 \quad (1)$ $(-1-h)^2 + (1-k)^2 = r^2 \quad (2)$ 2. The centre $(h,k)$ lies on the line $x-3y-11=0$, so: $h - 3k - 11 = 0 \implies h = 3k + 11 \quad (3)$ 3. Equating (1) and (2): $(2-h)^2 + (3-k)^2 = (-1-h)^2 + (1-k)^2$ $4-4h+h^2+9-6k+k^2 = 1+2h+h^2+1-2k+k^2$ $13 - 4h - 6k = 2 + 2h - 2k \implies 6h + 4k = 11 \quad (4)$ 4. Substitute $h$ from (3) into (4): $6(3k+11) + 4k = 11 \implies 18k + 66 + 4k = 11 \implies 22k = -55 \implies k = -5/2$. Substitute $k = -5/2$ into (3): $h = 3(-5/2) + 11 = -15/2 + 22/2 = 7/2$. So, the centre is $(7/2, -5/2)$. 5. Now, find $r^2$ using point $(2,3)$ and the centre: $r^2 = (2-7/2)^2 + (3-(-5/2))^2 = (-3/2)^2 + (11/2)^2 = 9/4 + 121/4 = 130/4 = 65/2$. The equation of the circle is: $(x-7/2)^2 + (y+5/2)^2 = 65/2$ $x^2 - 7x + 49/4 + y^2 + 5y + 25/4 = 65/2$ $x^2 + y^2 - 7x + 5y + 74/4 = 130/4$ $x^2 + y^2 - 7x + 5y - 56/4 = 0$ $x^2 + y^2 - 7x + 5y - 14 = 0$ The final answer is $\boxed{x^2 + y^2 - 7x + 5y - 14 = 0}$.

ID: 685 Type: Subjective Source: NCERT Class 11

Question 12: Find the equation of the circle with radius $5$ whose centre lies on $x$-axis and passes through the point $(2,3)$.

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🤖 AI Generated Solution:

The general equation of a circle with centre $(h,k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$. Given that the radius $r=5$ and the centre lies on the $x$-axis, the centre can be written as $(h,0)$. 1. Substitute the centre and radius into the general equation: $(x-h)^2 + (y-0)^2 = 5^2$ $(x-h)^2 + y^2 = 25$ 2. The circle passes through the point $(2,3)$. Substitute $x=2$ and $y=3$ into the equation to find $h$: $(2-h)^2 + 3^2 = 25$ $(2-h)^2 + 9 = 25$ $(2-h)^2 = 16$ 3. Solve for $h$: $2-h = \pm\sqrt{16}$ $2-h = \pm 4$ Case 1: $2-h = 4 \implies h = 2-4 \implies h = -2$. Case 2: $2-h = -4 \implies h = 2-(-4) \implies h = 6$. 4. Substitute the values of $h$ back into the circle's equation: For $h=-2$: $(x-(-2))^2 + y^2 = 25 \implies (x+2)^2 + y^2 = 25$ For $h=6$: $(x-6)^2 + y^2 = 25$ The equations of the circles are $(x+2)^2 + y^2 = 25$ and $(x-6)^2 + y^2 = 25$.

ID: 686 Type: Subjective Source: NCERT Class 11

Question 13: Find the equation of the circle passing through $(0,0)$ and making intercepts $a$ and $b$ on the coordinate axes.

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🤖 AI Generated Solution:

The general equation of a circle is given by $x^2 + y^2 + 2gx + 2fy + c = 0$. 1. **Circle passes through $(0,0)$**: Substitute $(0,0)$ into the general equation: $0^2 + 0^2 + 2g(0) + 2f(0) + c = 0$ This implies $c = 0$. So, the equation simplifies to $x^2 + y^2 + 2gx + 2fy = 0$. 2. **Circle makes an x-intercept $a$**: Since the circle passes through $(0,0)$ and makes an x-intercept of $a$, it means the circle intersects the x-axis at $(0,0)$ and $(a,0)$. Substitute the point $(a,0)$ into the simplified equation: $a^2 + 0^2 + 2g(a) + 2f(0) = 0$ $a^2 + 2ga = 0$ $a(a + 2g) = 0$ Since $a$ is an intercept, $a \neq 0$. Therefore, $a + 2g = 0$, which gives $2g = -a \implies g = -\frac{a}{2}$. 3. **Circle makes a y-intercept $b$**: Similarly, since the circle passes through $(0,0)$ and makes a y-intercept of $b$, it means the circle intersects the y-axis at $(0,0)$ and $(0,b)$. Substitute the point $(0,b)$ into the simplified equation: $0^2 + b^2 + 2g(0) + 2f(b) = 0$ $b^2 + 2fb = 0$ $b(b + 2f) = 0$ Since $b$ is an intercept, $b \neq 0$. Therefore, $b + 2f = 0$, which gives $2f = -b \implies f = -\frac{b}{2}$. 4. **Form the final equation**: Substitute the values of $g = -\frac{a}{2}$, $f = -\frac{b}{2}$, and $c = 0$ back into the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$: $x^2 + y^2 + 2\left(-\frac{a}{2}\right)x + 2\left(-\frac{b}{2}\right)y + 0 = 0$ $x^2 + y^2 - ax - by = 0$ The final answer is $\boxed{x^2+y^2-ax-by=0}$.

ID: 687 Type: Subjective Source: NCERT Class 11

Question 14: Find the equation of a circle with centre $(2,2)$ and passes through the point $(4,5)$.

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🤖 AI Generated Solution:

To find the equation of the circle, we use the standard form $(x-h)^2 + (y-k)^2 = r^2$, where $(h,k)$ is the centre and $r$ is the radius. 1. **Substitute the centre:** The given centre is $(h,k) = (2,2)$. So the equation becomes $(x-2)^2 + (y-2)^2 = r^2$. 2. **Find the radius squared ($r^2$):** The circle passes through the point $(4,5)$. We can substitute these coordinates into the equation to find $r^2$: $(4-2)^2 + (5-2)^2 = r^2$ $(2)^2 + (3)^2 = r^2$ $4 + 9 = r^2$ $13 = r^2$ 3. **Write the final equation:** Substitute the value of $r^2=13$ back into the equation from step 1: $(x-2)^2 + (y-2)^2 = 13$ The final answer is $\boxed{(x-2)^2 + (y-2)^2 = 13}$.

ID: 688 Type: Subjective Source: NCERT Class 11

Question 15: Does the point $(-2.5, 3.5)$ lie inside, outside or on the circle $x^{2}+y^{2}=25$?

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🤖 AI Generated Solution:

To determine the position of the point relative to the circle, substitute the coordinates of the point into the circle's equation and compare the result with the square of the radius. 1. The equation of the circle is $x^2 + y^2 = 25$. This means the circle is centered at $(0,0)$ and has a radius $r = \sqrt{25} = 5$. 2. Substitute the point $(-2.5, 3.5)$ into the left-hand side of the equation: $(-2.5)^2 + (3.5)^2$ 3. Calculate the value: $6.25 + 12.25 = 18.5$ 4. Compare this value with $r^2 = 25$. Since $18.5 < 25$, the distance from the origin to the point is less than the radius. Thus, the point $(-2.5, 3.5)$ lies inside the circle $x^{2}+y^{2}=25$. The final answer is $\boxed{\text{inside}}$.

ID: 689 Type: Subjective Source: NCERT Class 11

Question 16: Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for the parabola $y^{2}=12x$.

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🤖 AI Generated Solution:

The given equation of the parabola is $y^{2}=12x$. 1. **Standard Form Comparison:** This equation is in the standard form $y^{2}=4ax$. Comparing $y^{2}=12x$ with $y^{2}=4ax$, we get $4a=12$, which implies $a=3$. 2. **Coordinates of the Focus:** For a parabola of the form $y^{2}=4ax$, the focus is at $(a, 0)$. Substituting $a=3$, the focus is $(3, 0)$. 3. **Axis of the Parabola:** For a parabola of the form $y^{2}=4ax$, the axis is the x-axis. The equation of the axis is $y=0$. 4. **Equation of the Directrix:** For a parabola of the form $y^{2}=4ax$, the directrix is $x=-a$. Substituting $a=3$, the equation of the directrix is $x=-3$. 5. **Length of the Latus Rectum:** For a parabola of the form $y^{2}=4ax$, the length of the latus rectum is $4a$. Substituting $a=3$, the length of the latus rectum is $4(3)=12$. The final answer is $\boxed{\text{Focus: }(3,0), \text{ Axis: } y=0, \text{ Directrix: } x=-3, \text{ Latus Rectum: } 12}$.

ID: 690 Type: Subjective Source: NCERT Class 11

Question 17: Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for the parabola $x^{2}=6y$.

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🤖 AI Generated Solution:

The given equation of the parabola is $x^{2}=6y$. 1. **Standard Form Comparison:** The standard form of a parabola opening upwards is $x^2 = 4ay$. Comparing $x^{2}=6y$ with $x^2 = 4ay$, we have $4a = 6$, which implies $a = \frac{6}{4} = \frac{3}{2}$. 2. **Coordinates of the Focus:** For a parabola of the form $x^2 = 4ay$, the focus is at $(0, a)$. Focus: $(0, \frac{3}{2})$. 3. **Axis of the Parabola:** For a parabola of the form $x^2 = 4ay$, the axis of the parabola is the y-axis. Equation of the axis: $x=0$. 4. **Equation of the Directrix:** For a parabola of the form $x^2 = 4ay$, the equation of the directrix is $y = -a$. Equation of the directrix: $y = -\frac{3}{2}$. 5. **Length of the Latus Rectum:** The length of the latus rectum for a parabola of the form $x^2 = 4ay$ is $4a$. Length of the latus rectum: $4 \times \frac{3}{2} = 6$. The final answer is $\boxed{\text{Focus: }(0, \frac{3}{2}), \text{ Axis: } x=0, \text{ Directrix: } y=-\frac{3}{2}, \text{ Latus Rectum: } 6}$.

ID: 691 Type: Subjective Source: NCERT Class 11

Question 18: Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for the parabola $y^{2}=-8x$.

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🤖 AI Generated Solution:

The given equation of the parabola is $y^{2}=-8x$. 1. **Compare with the standard form:** This equation is of the form $y^{2}=-4ax$. Comparing $y^{2}=-4ax$ with $y^{2}=-8x$, we get $-4a = -8$, which implies $a=2$. 2. **Coordinates of the focus:** For a parabola of the form $y^{2}=-4ax$, the focus is at $(-a, 0)$. Substituting $a=2$, the focus is $(-2, 0)$. 3. **Axis of the parabola:** For a parabola of the form $y^{2}=-4ax$, the axis is the x-axis. The equation of the axis is $y=0$. 4. **Equation of the directrix:** For a parabola of the form $y^{2}=-4ax$, the equation of the directrix is $x=a$. Substituting $a=2$, the equation of the directrix is $x=2$. 5. **Length of the latus rectum:** The length of the latus rectum for any standard parabola is $4a$. Substituting $a=2$, the length of the latus rectum is $4(2)=8$. Thus, for the parabola $y^{2}=-8x$: * Focus: $(-2, 0)$ * Axis: $y=0$ * Directrix: $x=2$ * Length of latus rectum: $8$

ID: 692 Type: Subjective Source: NCERT Class 11

Question 19: Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for the parabola $x^{2}=-16y$.

ID: 693 Type: Subjective Source: NCERT Class 11

Question 20: Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for the parabola $y^{2}=10x$.

ID: 694 Type: Subjective Source: NCERT Class 11

Question 21: Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for the parabola $x^{2}=-9y$.

ID: 695 Type: Subjective Source: NCERT Class 11

Question 22: Find the equation of the parabola with Focus $(6,0)$ and directrix $x=-6$.

ID: 696 Type: Subjective Source: NCERT Class 11

Question 23: Find the equation of the parabola with Focus $(0,-3)$ and directrix $y=3$.

ID: 697 Type: Subjective Source: NCERT Class 11

Question 24: Find the equation of the parabola with Vertex $(0,0)$ and focus $(3,0)$.

ID: 698 Type: Subjective Source: NCERT Class 11

Question 25: Find the equation of the parabola with Vertex $(0,0)$ and focus $(-2,0)$.

ID: 699 Type: Subjective Source: NCERT Class 11

Question 26: Find the equation of the parabola with Vertex $(0,0)$, passing through $(2,3)$ and axis is along $x$-axis.

ID: 700 Type: Subjective Source: NCERT Class 11

Question 27: Find the equation of the parabola with Vertex $(0,0)$, passing through $(5,2)$ and symmetric with respect to $y$-axis.

ID: 701 Type: Subjective Source: NCERT Class 11

Question 28: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$.

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🤖 AI Generated Solution:

Here's how to find the properties of the given ellipse: **1. Identify the Ellipse Type and Parameters:** The given equation is $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$. This is an ellipse centered at the origin. Since $36 > 16$, the major axis is horizontal. We have $a^2 = 36$ and $b^2 = 16$. Therefore, $a = \sqrt{36} = 6$ and $b = \sqrt{16} = 4$. **2. Calculate the Distance from Center to Focus ($c$):** Use the relation $c^2 = a^2 - b^2$. $c^2 = 36 - 16 = 20$ $c = \sqrt{20} = 2\sqrt{5}$. **3. Determine the Foci:** Since the major axis is horizontal, the foci are at $(\pm c, 0)$. Foci: $(\pm 2\sqrt{5}, 0)$. **4. Determine the Vertices:** Since the major axis is horizontal, the vertices are at $(\pm a, 0)$. Vertices: $(\pm 6, 0)$. **5. Calculate the Length of the Major Axis:** The length of the major axis is $2a$. Length of Major Axis: $2 \times 6 = 12$. **6. Calculate the Length of the Minor Axis:** The length of the minor axis is $2b$. Length of Minor Axis: $2 \times 4 = 8$. **7. Calculate the Eccentricity ($e$):** Eccentricity is given by $e = \frac{c}{a}$. $e = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3}$. **8. Calculate the Length of the Latus Rectum:** The length of the latus rectum is $\frac{2b^2}{a}$. Length of Latus Rectum: $\frac{2 \times 16}{6} = \frac{32}{6} = \frac{16}{3}$. **Summary of Results:** * **Foci:** $(\pm 2\sqrt{5}, 0)$ * **Vertices:** $(\pm 6, 0)$ * **Length of Major Axis:** 12 * **Length of Minor Axis:** 8 * **Eccentricity:** $\frac{\sqrt{5}}{3}$ * **Length of Latus Rectum:** $\frac{16}{3}$

ID: 702 Type: Subjective Source: NCERT Class 11

Question 29: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{25}=1$.

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🤖 AI Generated Solution:

Here's the step-by-step solution for the given ellipse: The equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{25}=1$. Comparing this with the standard equation $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$ (since $25 > 4$, the major axis is along the y-axis), we have $a^2 = 25$ and $b^2 = 4$. 1. **Find the values of a and b:** $a = \sqrt{25} = 5$ $b = \sqrt{4} = 2$ 2. **Find the value of c:** For an ellipse, $c^2 = a^2 - b^2$. $c^2 = 25 - 4 = 21$ $c = \sqrt{21}$ 3. **Coordinates of the foci:** Since the major axis is along the y-axis, the foci are at $(0, \pm c)$. Foci: $(0, \sqrt{21})$ and $(0, -\sqrt{21})$. 4. **Coordinates of the vertices:** The vertices are at $(0, \pm a)$. Vertices: $(0, 5)$ and $(0, -5)$. 5. **Length of the major axis:** Length of major axis = $2a = 2 \times 5 = 10$. 6. **Length of the minor axis:** Length of minor axis = $2b = 2 \times 2 = 4$. 7. **Eccentricity (e):** $e = \frac{c}{a} = \frac{\sqrt{21}}{5}$. 8. **Length of the latus rectum:** Length of latus rectum = $\frac{2b^2}{a} = \frac{2 \times 4}{5} = \frac{8}{5}$. The final answer is $\boxed{Foci: (0, \pm \sqrt{21}), Vertices: (0, \pm 5), Length of major axis: 10, Length of minor axis: 4, Eccentricity: \frac{\sqrt{21}}{5}, Length of latus rectum: \frac{8}{5}}$.

ID: 703 Type: Subjective Source: NCERT Class 11

Question 30: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$.

ID: 704 Type: Subjective Source: NCERT Class 11

Question 31: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{100}=1$.

ID: 705 Type: Subjective Source: NCERT Class 11

Question 32: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{x^{2}}{49}+\frac{y^{2}}{36}=1$.

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🤖 AI Generated Solution:

The given equation of the ellipse is $\frac{x^{2}}{49}+\frac{y^{2}}{36}=1$. **1. Identify the semi-major and semi-minor axes:** Comparing with the standard form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we have $a^2 = 49$ and $b^2 = 36$. So, $a = \sqrt{49} = 7$ and $b = \sqrt{36} = 6$. Since $a > b$, the major axis is along the x-axis. **2. Calculate the distance from the center to the foci ($c$):** We use the relation $c^2 = a^2 - b^2$. $c^2 = 49 - 36 = 13$ $c = \sqrt{13}$. **3. Determine the coordinates of the foci:** The foci are at $(\pm c, 0)$. Foci are $(\sqrt{13}, 0)$ and $(-\sqrt{13}, 0)$. **4. Determine the coordinates of the vertices:** The vertices are at $(\pm a, 0)$. Vertices are $(7, 0)$ and $(-7, 0)$. **5. Calculate the length of the major axis:** The length of the major axis is $2a$. Length of major axis = $2 \times 7 = 14$. **6. Calculate the length of the minor axis:** The length of the minor axis is $2b$. Length of minor axis = $2 \times 6 = 12$. **7. Calculate the eccentricity ($e$):** The eccentricity is given by $e = \frac{c}{a}$. $e = \frac{\sqrt{13}}{7}$. **8. Calculate the length of the latus rectum:** The length of the latus rectum is $\frac{2b^2}{a}$. Length of latus rectum = $\frac{2 \times 36}{7} = \frac{72}{7}$. The final answer is $\boxed{Foci: (\pm \sqrt{13}, 0), Vertices: (\pm 7, 0), Length of major axis: 14, Length of minor axis: 12, Eccentricity: \frac{\sqrt{13}}{7}, Length of latus rectum: \frac{72}{7}}$.

ID: 706 Type: Subjective Source: NCERT Class 11

Question 33: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{x^{2}}{100}+\frac{y^{2}}{400}=1$.

ID: 707 Type: Subjective Source: NCERT Class 11

Question 34: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $36x^{2}+4y^{2}=144$.

ID: 708 Type: Subjective Source: NCERT Class 11

Question 35: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $16x^{2}+y^{2}=16$.

ID: 709 Type: Subjective Source: NCERT Class 11

Question 36: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $4x^{2}+9y^{2}=36$.

ID: 710 Type: Subjective Source: NCERT Class 11

Question 37: Find the equation for the ellipse with Vertices $(\pm 5, 0)$, foci $(\pm 4, 0)$.

ID: 711 Type: Subjective Source: NCERT Class 11

Question 38: Find the equation for the ellipse with Vertices $(0, \pm 13)$, foci $(0, \pm 5)$.

ID: 712 Type: Subjective Source: NCERT Class 11

Question 39: Find the equation for the ellipse with Vertices $(\pm 6, 0)$, foci $(\pm 4, 0)$.

ID: 713 Type: Subjective Source: NCERT Class 11

Question 40: Find the equation for the ellipse with ends of major axis $(\pm 3, 0)$, ends of minor axis $(0, \pm 2)$.

ID: 714 Type: Subjective Source: NCERT Class 11

Question 41: Find the equation for the ellipse with ends of major axis $(0, \pm \sqrt{5})$, ends of minor axis $(\pm 1, 0)$.

ID: 715 Type: Subjective Source: NCERT Class 11

Question 42: Find the equation for the ellipse with length of major axis $26$, foci $(\pm 5, 0)$.

ID: 716 Type: Subjective Source: NCERT Class 11

Question 43: Find the equation for the ellipse with length of minor axis $16$, foci $(0, \pm 6)$.

ID: 717 Type: Subjective Source: NCERT Class 11

Question 44: Find the equation for the ellipse with Foci $(\pm 3, 0)$, $a=4$.

ID: 718 Type: Subjective Source: NCERT Class 11

Question 45: Find the equation for the ellipse with $b=3$, $c=4$, centre at the origin; foci on the $x$ axis.

ID: 719 Type: Subjective Source: NCERT Class 11

Question 46: Find the equation for the ellipse with Centre at $(0,0)$, major axis on the $y$-axis and passes through the points $(3, 2)$ and $(1,6)$.

ID: 720 Type: Subjective Source: NCERT Class 11

Question 47: Find the equation for the ellipse with Major axis on the $x$-axis and passes through the points $(4,3)$ and $(6,2)$.

ID: 721 Type: Subjective Source: NCERT Class 11

Question 48: Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$.

ID: 722 Type: Subjective Source: NCERT Class 11

Question 49: Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola $\frac{y^{2}}{9}-\frac{x^{2}}{27}=1$.

ID: 723 Type: Subjective Source: NCERT Class 11

Question 50: Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola $9y^{2}-4x^{2}=36$.

ID: 724 Type: Subjective Source: NCERT Class 11

Question 51: Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola $16x^{2}-9y^{2}=576$.

ID: 725 Type: Subjective Source: NCERT Class 11

Question 52: Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola $5y^{2}-9x^{2}=36$.

ID: 726 Type: Subjective Source: NCERT Class 11

Question 53: Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola $49y^{2}-16x^{2}=784$.

ID: 727 Type: Subjective Source: NCERT Class 11

Question 54: Find the equation of the hyperbola with Vertices $(\pm 2, 0)$, foci $(\pm 3, 0)$.

ID: 728 Type: Subjective Source: NCERT Class 11

Question 55: Find the equation of the hyperbola with Vertices $(0, \pm 5)$, foci $(0, \pm 8)$.

ID: 729 Type: Subjective Source: NCERT Class 11

Question 56: Find the equation of the hyperbola with Vertices $(0, \pm 3)$, foci $(0, \pm 5)$.

ID: 730 Type: Subjective Source: NCERT Class 11

Question 57: Find the equation of the hyperbola with Foci $(\pm 5, 0)$, the transverse axis is of length $8$.

ID: 731 Type: Subjective Source: NCERT Class 11

Question 58: Find the equation of the hyperbola with Foci $(0, \pm 13)$, the conjugate axis is of length $24$.

ID: 732 Type: Subjective Source: NCERT Class 11

Question 59: Find the equation of the hyperbola with Foci $(\pm 3\sqrt{5}, 0)$, the latus rectum is of length $8$.

ID: 733 Type: Subjective Source: NCERT Class 11

Question 60: Find the equation of the hyperbola with Foci $(\pm 4, 0)$, the latus rectum is of length $12$.

ID: 734 Type: Subjective Source: NCERT Class 11

Question 61: Find the equation of the hyperbola with vertices $(\pm 7,0)$, $e=\frac{4}{3}$.

ID: 735 Type: Subjective Source: NCERT Class 11

Question 62: Find the equation of the hyperbola with Foci $(0, \pm \sqrt{10})$, passing through $(2,3)$.

ID: 736 Type: Subjective Source: NCERT Class 11

Question 63: If a parabolic reflector is $20$ cm in diameter and $5$ cm deep, find the focus.

ID: 737 Type: Subjective Source: NCERT Class 11

Question 64: An arch is in the form of a parabola with its axis vertical. The arch is $10$ m high and $5$ m wide at the base. How wide is it $2$ m from the vertex of the parabola?

ID: 738 Type: Subjective Source: NCERT Class 11

Question 65: The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and $100$ m long is supported by vertical wires attached to the cable, the longest wire being $30$ m and the shortest being $6$ m. Find the length of a supporting wire attached to the roadway $18$ m from the middle.

ID: 739 Type: Subjective Source: NCERT Class 11

Question 66: An arch is in the form of a semi-ellipse. It is $8$ m wide and $2$ m high at the centre. Find the height of the arch at a point $1.5$ m from one end.

ID: 740 Type: Subjective Source: NCERT Class 11

Question 67: A rod of length $12$ cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point $P$ on the rod, which is $3$ cm from the end in contact with the $x$-axis.

ID: 741 Type: Subjective Source: NCERT Class 11

Question 68: Find the area of the triangle formed by the lines joining the vertex of the parabola $x^{2}=12y$ to the ends of its latus rectum.

ID: 742 Type: Subjective Source: NCERT Class 11

Question 69: A man running a racecourse notes that the sum of the distances from the two flag posts from him is always $10$ m and the distance between the flag posts is $8$ m. Find the equation of the posts traced by the man.

ID: 743 Type: Subjective Source: NCERT Class 11

Question 70: An equilateral triangle is inscribed in the parabola $y^{2}=4ax$, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

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🤖 AI Generated Solution:

Let the parabola be $y^2 = 4ax$. Its vertex is at the origin $O(0,0)$. Let $O$ be one vertex of the equilateral triangle. Due to the symmetry of the parabola about the x-axis, the other two vertices, say $A$ and $B$, must be symmetric with respect to the x-axis. Let $A = (x,y)$ and $B = (x,-y)$. Since $\triangle OAB$ is equilateral and $O$ is at the origin, the line segments $OA$ and $OB$ must make angles of $30^\circ$ and $-30^\circ$ with the positive x-axis, respectively. For vertex $A(x,y)$, we have $\frac{y}{x} = \tan(30^\circ) = \frac{1}{\sqrt{3}}$. So, $y = \frac{x}{\sqrt{3}}$. Substitute this into the equation of the parabola $y^2 = 4ax$: $$ \left(\frac{x}{\sqrt{3}}\right)^2 = 4ax $$ $$ \frac{x^2}{3} = 4ax $$ Since $x \neq 0$ (otherwise $A$ would coincide with $O$, forming a degenerate triangle), we can divide by $x$: $$ \frac{x}{3} = 4a \implies x = 12a $$ Now find $y$: $$ y = \frac{12a}{\sqrt{3}} = 4\sqrt{3}a $$ So the coordinates of vertex $A$ are $(12a, 4\sqrt{3}a)$. The length of the side of the triangle $s$ is the distance $OA$: $$ s = \sqrt{(x-0)^2 + (y-0)^2} = \sqrt{x^2 + y^2} $$ $$ s = \sqrt{(12a)^2 + (4\sqrt{3}a)^2} = \sqrt{144a^2 + (16 \times 3)a^2} $$ $$ s = \sqrt{144a^2 + 48a^2} = \sqrt{192a^2} = \sqrt{64 \times 3 a^2} $$ $$ s = 8\sqrt{3}a $$ The final answer is $\boxed{8\sqrt{3}a}$.