Conic Sections

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The length of the latus rectum of the ellipse $3 x^{2} + y^{2} = 12$ is

(a) 4
(b) 3
(c) $\frac{8}{3}$
(d) $\frac{4}{3}$

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} E q u a t i o n o f t h e e l l i p s e i s 3 x^{2} + y^{2} = 12 \\ \Rightarrow \frac{x^{2}}{4} + \frac{y^{2}}{12} = 1 \\ H e r e a^{2} = 4 \Rightarrow a = 2 \\ b^{2} = 12 \Rightarrow b = 2 \sqrt[]{3} \\ L e n g t h o f t h e l a t u s r e c t u m = \frac{2 a^{2}}{b} = \frac{2 \times 4}{2 \sqrt[]{3}} = \frac{4}{\sqrt[]{3}} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The equation of the ellipse whose focus is (1, â€“1), the directrix is the line $x - y - 3 = 0$ , and the eccentricity is $\frac{1}{2}$ is

(a) $7 x^{2} + 2 x y + 7 y^{2} - 10 x + 10 y + 7 = 0$
(b) $7 x^{2} + 2 x y + 7 y^{2} + 7 = 0$
(c) $7 x^{2} + 2 x y + 7 y^{2} + 10 x - 10 y - 7 = 0$
(d) None of these

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f o c u s o f t h e e l l i p s e i s (1, - 1) a n d t h e e q u a t i o n o f t h e d i r e c t r i x i s x - y - 3 = 0 \\ a n d e = \frac{1}{2} . \\ L e t P (x, y) b e a n y point o n t h e p a r a b o l a \\ ∴ \frac{P F}{Distance o f t h e point P f r o m t h e d i r e c t r i x} = e \\ \Rightarrow \frac{\sqrt[]{{(x - 1)}^{2} + {(y + 1)}^{2}}}{| \frac{x - y - 3}{\sqrt[]{{(1)}^{2} + {(- 1)}^{2}}} |} = \frac{1}{2} \\ \Rightarrow 2 \sqrt[]{x^{2} + 1 - 2 x + y^{2} + 1 + 2 y} = | \frac{x - y - 3}{\sqrt[]{2}} | \\ S q u a r i n g b o t h s i d e s, w e h a v e \\ \Rightarrow 4 (x^{2} + y^{2} - 2 x + 2 y + 2) = \frac{x^{2} + y^{2} + 9 - 2 x y + 6 y - 6 x}{2} \\ \Rightarrow 8 x^{2} + 8 y^{2} - 16 x + 16 y + 16 = x^{2} + y^{2} + 9 - 2 x y + 6 y - 6 x \\ \Rightarrow 7 x^{2} + 7 y^{2} + 2 x y - 10 x + 10 y + 7 = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

If the vertex of the parabola is at the point (â€“3, 0) and the directrix is the line $x + 5 = 0$ , then its equation is

(a) $y^{2} = - 8 (x + 3)$
(b) $x^{2} = 8 (y + 3)$
(c) $y^{2} = 8 (x + 3)$
(d) $y^{2} = 8 (x + 5)$

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t v e r t e x = (- 3, 0) \\ ∴ a = - 3 a n d d i r e c t r i x i s x + 5 = 0 \\ A c c o r d i n g t o t h e d e f i n i t i o n o f t h e p a r a b o l a, w e g e t \\ A F = A D i . e ., A i s t h e m i d point o f D F \\ ∴ - 3 = \frac{x_{1} - 5}{2} \Rightarrow x_{1} = - 6 + 5 = - 1 \\ a n d 0 = \frac{0 + y_{1}}{2} \Rightarrow y_{1} = 0 \\ ∴ F o c u s F = (- 1, 0) \\ N o w \sqrt[]{{(x + 1)}^{2} + {(y - 0)}^{2}} = | \frac{x + 5}{\sqrt[]{1^{2} + 0^{2}}} | \\ S q u a r i n g b o t h s i d e s, w e g e t \\ {(x + 1)}^{2} + {(y - 0)}^{2} = {(x + 5)}^{2} \\ \Rightarrow x^{2} + 1 + 2 x + y^{2} = x^{2} + 25 + 10 x \\ \Rightarrow y^{2} = 10 x - 2 x + 24 \Rightarrow y^{2} = 8 x + 24 \\ \Rightarrow y^{2} = 8 (x + 3) \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

If the parabola $y^{2} = 4 a x$ passes through the point (3, 2), then the length of its latus rectum is

(a) $\frac{2}{3}$
(b) 2
(c) 4
(d) None of these

11 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$G i v e n p a r a b o l a i s y^{2} = 4 a x I f t h e p a r a b o l a i s passing$
$\begin{array}{l} t h r o u g h (3, 2) \\ t h e n {(2)}^{2} = 4 a \times 3 \\ \Rightarrow 4 = 12 a \Rightarrow a = \frac{1}{3} \\ N o w l e n g t h o f t h e l a t u s r e c t u m = 4 a = 4 \times \frac{1}{3} = \frac{4}{3} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

If the focus of a parabola is (0, -3) and its directrix is $y = 3$ , then its equation is

(a) $x^{2} = - 12 y$
(b) $x^{2} = 12 y$
(c) $y^{2} = - 12 x$
(d) $y^{2} = 12 x$

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} A c c o r d i n g t o t h e d e f i n i t i o n o f p a r a b o l a \\ \sqrt[]{{(x - 0)}^{2} + {(y + 3)}^{2}} = | \frac{y - 3}{\sqrt[]{{(0)}^{2} + {(1)}^{2}}} | \\ \Rightarrow \sqrt[]{x^{2} + y^{2} + 9 + 6 y} = | y - 3 | \\ S q u a r i n g b o t h s i d e s, w e h a v e \\ x^{2} + y^{2} + 9 + 6 y = y^{2} + 9 - 6 y \\ \Rightarrow x^{2} + 9 + 6 y = 9 - 6 y \\ \Rightarrow x^{2} = - 12 y \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The equation of a circle with the origin as the center and passing through the vertices of an equilateral triangle whose median is of length $3 a$ is

$(a)$ (a) $x^{2} + y^{2} = 9 a^{2}$
(b) $x^{2} + y^{2} = 9 a^{2}$
(c) $x^{2} + y^{2} = 4 a^{2}$
(d) $x^{2} + y^{2} = a^{2}$

[Hint: The centroid of the triangle coincides with the center of the circle, and the radius is $\frac{2}{3}$ of the median.]

8 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t A B C b e a n e q u i l a t e r a l t r i a n g l e i n w h i c h m e d i a n A D = 3 a . \\ C e n t r e o f t h e c i r c l e i s s a m e a s t h e c e n t r o i d o f t h e t r i a n g l e i . e ., (0, 0) \\ A G : G D = 2 : 1 \\ S o, A G = \frac{2}{3} A D = \frac{2}{3} \times 3 a = 2 a \\ ∴ T h e e q u a t i o n o f t h e c i r c l e i s \\ {(x - 0)}^{2} + {(y - 0)}^{2} = {(2 a)}^{2} \\ \Rightarrow x^{2} + y^{2} = 4 a^{2} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The equation of the circle with its center on the y-axis and passing through the origin and the point (2, 3) is

(a) $x^{2} + y^{2} + 13 y = 0$
(b) $x^{2} + y^{2} + 13 x + 3 = 0$
(c) $x^{2} + y^{2} - 13 y = 0$
(d) None of these

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t t h e e q u a t i o n o f t h e c i r c l e b e \\ {(x - h)}^{2} + {(y - k)}^{2} = r^{2} \\ L e t t h e c e n t r e b e (0, a) a n d r = a \\ S o, t h e e q u a t i o n o f t h e c i r c l e i s \\ {(x - 0)}^{2} + {(y - a)}^{2} = a^{2} \\ \Rightarrow x^{2} + {(y - a)}^{2} = a^{2} \\ \Rightarrow x^{2} + y^{2} + a^{2} - 2 a y = a^{2} \\ \Rightarrow x^{2} + y^{2} - 2 a y = 0 \dots (i) \\ N o w C P = r \\ \Rightarrow \sqrt[]{{(2 - 0)}^{2} + {(3 - a)}^{2}} = a \\ \Rightarrow \sqrt[]{4 + 9 + a^{2} - 6 a} = a \\ \Rightarrow \sqrt[]{13 + a^{2} - 6 a} = a \\ \Rightarrow 13 + a^{2} - 6 a = a^{2} \\ \Rightarrow 13 - 6 a = 0 \\ \Rightarrow a = \frac{13}{6} \\ P u t t i n g t h e v a l u e o f a i n e q n . (i) w e g e t \\ \Rightarrow x^{2} + y^{2} - 2 (\frac{13}{6}) y = 0 \\ \Rightarrow 3 x^{2} + 3 y^{2} - 13 y = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The equation of a circle that passes through (3, 6) and touches the axes is

(a) $x^{2} + y^{2} + 6 x + 6 y + 3 = 0$
(b) $x^{2} + y^{2} - 6 x - 6 y - 9 = 0$
(c) $x^{2} + y^{2} - 6 x - 6 y + 9 = 0$
(d) None of these

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t t h e r e q u i r e d c i r c l e t o u c h t h e a x e s a t (a, 0) a n d (0, a) \\ ∴ C e n t r e i s (a, a) a n d r = a \\ S o, t h e e q u a t i o n o f t h e c i r c l e i s \\ {(x - a)}^{2} + {(y - a)}^{2} = a^{2} \\ I f i t p a s s e s t h r o u g h a point P (3, 6) t h e n \\ {(3 - a)}^{2} + {(6 - a)}^{2} = a^{2} \\ \Rightarrow 9 + a^{2} - 6 a + 36 + a^{2} - 12 a = a^{2} \\ \Rightarrow a^{2} - 18 a + 45 = 0 \\ \Rightarrow a^{2} - 15 a - 3 a + 45 = 0 \\ \Rightarrow a (a - 15) - 3 (a - 15) = 0 \\ \Rightarrow (a - 3) (a - 15) = 0 \\ \Rightarrow a = 3 a n d a = 15 w h i c h i s n o t p o s s i b l e \\ ∴ a = 3 \\ S o, t h e r e q u i r e d e q u a t i o n o f t h e c i r c l e i s \\ {(x - 3)}^{2} + {(y - 3)}^{2} = 9 \\ \Rightarrow x^{2} + 9 - 6 x + y^{2} + 9 - 6 y = 9 \\ \Rightarrow x^{2} + y^{2} - 6 x - 6 y + 9 = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The area of the circle centered at (1, 2) and passing through (4, 6) is

(a) $5 π$

(b) $10 π$

(c) $25 π$

(d) None of these

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t t h e c e n t r e o f t h e c i r c l e i s (1, 2) \\ R a d i u s o f t h e c i r c l e = \sqrt[]{{(4 - 1)}^{2} + {(6 - 2)}^{2}} \\ = \sqrt[]{9 + 16} = 5 \\ S o, t h e a r e a o f c i r c l e = π r^{2} = π {(5)}^{2} = 25 π \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The equation of the hyperbola with vertices at $(0, \pm 6)$ , eccentricity $\frac{5}{3}$ , and foci at $(0, \pm 10)$ is ______

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t e q u a t i o n o f t h e h y p e r b o l a i s - \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \\ V e r t i c e s a r e (0, \pm b) ∴ b = 6 a n d e = \frac{5}{3} \\ W e k n o w t h a t e = \sqrt[]{1 + \frac{a^{2}}{b^{2}}} \\ \frac{5}{3} = \sqrt[]{1 + \frac{a^{2}}{36}} \Rightarrow \frac{25}{9} = 1 + \frac{a^{2}}{36} \\ \Rightarrow \frac{a^{2}}{36} = \frac{25}{9} - 1 = \frac{16}{9} \Rightarrow a^{2} = \frac{16}{9} \times 36 \\ \Rightarrow a^{2} = 64 \\ S o, t h e e q u a t i o n o f t h e h y p e r b o l a i s \\ - \frac{x^{2}}{64} + \frac{y^{2}}{36} = 1 \Rightarrow \frac{y^{2}}{36} - \frac{x^{2}}{64} = 1 \\ a n d f o c i = (0, \pm b e) = (0, \pm 6 \times \frac{5}{3}) = (0, \pm 10) \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s \frac{y^{2}}{36} - \frac{x^{2}}{64} = 1 a n d (0, \pm 10) . \end{array}$

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