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2 months ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the distance between the directrices of the ellipse $\frac{x^{2}}{36} + \frac{y^{2}}{20} = 1$ .

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n o f a n e l l i p s e i s \frac{x^{2}}{36} + \frac{y^{2}}{20} = 1 \\ H e r e, a^{2} = 36 \Rightarrow a = 6 a n d b^{2} = 20 \Rightarrow b = 2 \sqrt[]{5} \\ W e k n o w t h a t b^{2} = a^{2} (1 - e^{2}) \\ \Rightarrow 20 = 36 (1 - e^{2}) \\ \Rightarrow 1 - e^{2} = \frac{20}{36} \Rightarrow e^{2} = 1 - \frac{20}{36} = \frac{16}{36} \\ \Rightarrow e = \frac{4}{6} = \frac{2}{3} \\ N o w Distance b e t w e e n t h e d i r e c t r i c e s i s \\ \frac{a}{e} - (- \frac{a}{e}) = \frac{a}{e} + \frac{a}{e} = \frac{2 a}{e} \\ = 2 * \frac{6}{2 / 3} = 2 * 6 * \frac{3}{2} = 18 \\ H e n c e, t h e r e q u i r e d distance = 18 . \end{array}$

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New answer posted

2 months ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of the ellipse whose eccentricity is $\frac{\sqrt[]{3}}{2}$ , latus rectum is 5, and the center is (0, 0)

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} E q u a t i o n o f a n e l l i p s e i s \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \dots (i) \\ G i v e n t h a t, e = \frac{2}{3} \\ a n d l a t u s r e c t u m = \frac{2 b^{2}}{a} = 5 \\ \Rightarrow b^{2} = \frac{5}{2} a \dots (i i) \\ W e k n o w t h a t b^{2} = a^{2} (1 - e^{2}) \\ \Rightarrow \frac{5}{2} a = a^{2} (1 - \frac{4}{9}) \\ \Rightarrow \frac{5}{2} = a * \frac{5}{9} \Rightarrow a = \frac{9}{2} \Rightarrow a^{2} = \frac{81}{4} \\ a n d b^{2} = \frac{5}{2} * \frac{9}{2} = \frac{45}{4} \\ H e n c e, t h e r e q u i r e d e q u a t i o n o f e l l i p s e i s \\ \frac{x^{2}}{81 / 4} + \frac{y^{2}}{45 / 4} = 1 \Rightarrow \frac{4}{81} x^{2} + \frac{4}{45} y^{2} = 1 \end{array}$

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New answer posted

2 months ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

If the eccentricity of an ellipse is $\frac{5}{8}$ and the distance between its foci is 10, then find the latus rectum of the ellipse.

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} E q u a t i o n o f a n e l l i p s e i s \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \\ Distance b e t w e e n i t s f o c i = a e + a e = 2 a e \\ ∴ 2 a e = 10 \\ \Rightarrow a e = 5 \Rightarrow a * \frac{5}{8} = 5 \Rightarrow a = 8 \\ N o w b^{2} = a^{2} (1 - e^{2}), w h e r e e i s t h e e c c e n t r i c i t y \\ \Rightarrow b^{2} = 64 (1 - \frac{25}{64}) \\ \Rightarrow b^{2} = 64 * \frac{39}{64} \Rightarrow b^{2} = 39 \\ S o, t h e l e n g t h o f t h e l a t u s r e c t u m = \frac{2 b^{2}}{a} = \frac{2 * 39}{8} = \frac{39}{4} \\ H e n c e, l e n g t h o f t h e l a t u s r e c t u m = \frac{39}{4} . \end{array}$

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2 months ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Given the ellipse with equation $9 x^{2} + 25 y^{2} = 225$ , find the eccentricity and foci.

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n o f a n e l l i p s e i s \\ 9 x^{2} + 25 y^{2} = 225 \\ \Rightarrow \frac{9}{225} x^{2} + \frac{25}{225} y^{2} = 1 \\ \Rightarrow \frac{x^{2}}{25} + \frac{y^{2}}{9} = 1 \\ H e r e, a = 5 a n d b = 3 \\ N o w b^{2} = a^{2} (1 - e^{2}), w h e r e e i s t h e e c c e n t r i c i t y \\ \Rightarrow 9 = 25 (1 - e^{2}) \\ \Rightarrow 1 - e^{2} = \frac{9}{25} \Rightarrow e^{2} = 1 - \frac{9}{25} = \frac{16}{25} \\ ∴ e = \frac{4}{5} \\ N o w f o c i = (\pm a e, 0) = (\pm 5 * \frac{4}{5}, 0) = (\pm 4, 0) \\ H e n c e, e c c e n t r i c i t y i s \frac{4}{5}, f o c i = (\pm 4, 0) . \end{array}$

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New answer posted

2 months ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

If the latus rectum of an ellipse is equal to half of its minor axis, then find its eccentricity.

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t t h e e q u a t i o n o f a n e l l i p s e i s \\ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \\ L e n g t h o f m a j o r a x i s = 2 a \\ L e n g t h o f minor a x i s = 2 b \\ a n d t h e l e n g t h o f l a t u s r e c t u m = \frac{2 b^{2}}{a} \\ A c c o r d i n g t o t h e q u e s t i o n, w e h a v e \\ \frac{2 b^{2}}{a} = \frac{2 b}{2} \Rightarrow b = \frac{a}{2} \\ N o w b^{2} = a^{2} (1 - e^{2}), w h e r e e i s t h e e c c e n t r i c i t y \\ \Rightarrow b^{2} = 4 b^{2} (1 - e^{2}) \\ \Rightarrow 1 = 4 (1 - e^{2}) \Rightarrow 1 - e^{2} = \frac{1}{4} \Rightarrow e^{2} = \frac{3}{4} \\ ∴ e = \pm \frac{\sqrt[]{3}}{2} \\ S o, e = \frac{\sqrt[]{3}}{2} [? e i s n o t (-)] \\ H e n c e, t h e r e q u i r e d v a l u e o f e c c e n t r i c i t y i s \frac{\sqrt[]{3}}{2} . \end{array}$

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New answer posted

2 months ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of a circle concentric with the circle $x^{2} + y^{2} - 6 x + 12 y + 15 = 0$ and has double its area.

[Hint: Concentric circles have the same center.]

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n o f t h e c i r c l e i s \\ x^{2} + y^{2} - 6 x + 12 y + 15 = 0 \dots (i) \\ C e n t r e = (- g, - f) = (3, - 6) [\begin{array}{l} ? 2 g = - 6 \Rightarrow g = - 3 \\ 2 f = 12 \Rightarrow f = 6 \end{array}] \\ Since t h e c i r c l e i s c o n c e n t r i c w i t h t h e g i v e n c i r c l e \\ ∴ C e n t r e = (3, - 6) \\ N o w l e t t h e r a d i u s o f t h e c i r c l e i s r \\ ∴ r = \sqrt[]{g^{2} + f^{2} - c} = \sqrt[]{9 + 36 - 15} = \sqrt[]{30} \\ A r e a o f t h e g i v e n c i r c l e (i) = π r^{2} = 30 π s q . u n i t \\ A r e a o f t h e r e q u i r e d c i r c l e = 2 * 30 π = 60 π s q . u n i t \\ I f r_{1} b e t h e r a d i u s o f t h e r e q u i r e d c i r c l e \\ π r_{1}^{2} = 60 π \Rightarrow r_{1}^{2} = 60 \\ S o, t h e r e q u i r e d e q u a t i o n o f t h e c i r c l e i s \\ {(x - 3)}^{2} + {(y + 6)}^{2} = 60 \\ \Rightarrow x^{2} + 9 - 6 x + y^{2} + 36 + 12 y - 60 = 0 \\ \Rightarrow x^{2} + y^{2} - 6 x + 12 y - 15 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} + y^{2} - 6 x + 12 y - 15 = 0 . \end{array}$

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New answer posted

2 months ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

If the line $y = \sqrt[]{3} x + k$ touches the circle $x^{2} + y^{2} = 16$ , then find the value of $k$ .

[Hint: Equate perpendicular distance from the center of the circle to its radius.]

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

2 months ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of the circle having $(1, - 2)$ as its center and passing through the line $3 x + y = 14$ , $2 x + 5 y = 18$ .

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n s a r e \\ 3 x + y = 14 \dots (i) \\ a n d 2 x + 5 y = 18 \dots (i i) \\ F r o m e q . (i) w e g e t \\ y = 14 - 3 x \dots (i i i) \\ P u t t i n g t h e v a l u e o f y i n e q . (i i) w e g e t \\ \Rightarrow 2 x + 5 (14 - 3 x) = 18 \\ \Rightarrow 2 x + 70 - 15 x = 18 \\ \Rightarrow - 13 x = - 70 + 18 \\ \Rightarrow - 13 x = - 52 \Rightarrow x = 4 \\ F r o m e q . (i i i) w e g e t, y = 14 - 3 * 4 = 2 \\ ∴ point o fintersection i s (4, 2) \\ N o w, r a d i u s r = \sqrt[]{{(4 - 1)}^{2} + {(2 + 2)}^{2}} = \sqrt[]{{(3)}^{2} + {(4)}^{2}} = \sqrt[]{9 + 16} = 5 \\ S o, t h e e q u a t i o n o f c i r c l e i s \\ {(x - h)}^{2} + {(y - k)}^{2} = r^{2} \\ \Rightarrow {(x - 1)}^{2} + {(y + 2)}^{2} = {(5)}^{2} \\ \Rightarrow x^{2} - 2 x + 1 + y^{2} + 4 y + 4 = 25 \\ \Rightarrow x^{2} + y^{2} - 2 x + 4 y - 20 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} + y^{2} - 2 x + 4 y - 20 = 0 . \end{array}$

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New answer posted

2 months ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

If one end of a diameter of the circle $x^{2} + y^{2} - 4 x - 6 y + 11 = 0$ is $(3, 4)$ , then find the coordinates of the other end of the diameter.

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t t h e o t h e r e n d o f t h e d i a m e t e r i s (x_{1}, y_{1}) . \\ E q u a t i o n o f t h e g i v e n c i r c l e i s \\ x^{2} + y^{2} - 4 x - 6 y + 11 = 0 \\ C e n t r e = (- g, - f) = (2, 3) \\ ∴ \frac{x_{1} + 3}{2} = 2 \Rightarrow x_{1} + 3 = 4 \Rightarrow x_{1} = 1 \\ a n d \frac{y_{1} + 4}{2} = 3 \Rightarrow y_{1} + 4 = 6 \Rightarrow y_{1} = 2 \\ H e n c e, t h e r e q u i r e d c o o r d i n a t e s a r e (1, 2) . \end{array}$

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New answer posted

2 months ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of a circle which touches both axes and the line $3 x - 4 y + 8 = 0$ and lies in the third quadrant.
[Hint: Let $a$ be the radius of the circle, then $(- a, - a)$ will be the center, and the perpendicular distance from the center to the given line gives the radius of the circle.]

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

Sol:

$\begin{array}{l} L e t a b e t h e r a d i u s o f t h e c i r c l e . \\ C e n t r e o f t h e c i r c l e = (- a, - a) \\ Distance o f t h e l i n e 3 x - 4 y + 8 = 0 \\ F r o m t h e c e n t r e = R a d i u s o f t h e c i r c l e \\ | \frac{- 3 a + 4 a + 8}{\sqrt[]{{(3)}^{2} + {(- 4)}^{2}}} | = a \\ \Rightarrow | \frac{a + 8}{5} | = \frac{3}{2} \\ \Rightarrow \pm (\frac{a + 8}{5}) = a \\ \Rightarrow \frac{a + 8}{5} = a a n d - (\frac{a + 8}{5}) = a \\ \Rightarrow a = 5 a - 8 \\ \Rightarrow 4 a = 8 \Rightarrow a = 2 \\ a n d \frac{a + 8}{5} = - a \Rightarrow a + 8 = - 5 a \\ \Rightarrow 6 a = - 8 \Rightarrow a = - \frac{4}{3} \\ ∴ a = 2 a n d a \neq - \frac{4}{3} \\ ∴ T h e e q u a t i o n o f t h e c i r c l e i s \\ {(x + 2)}^{2} + {(y + 2)}^{2} = {(2)}^{2} \\ \Rightarrow x^{2} + 4 x + 4 + y^{2} + 4 y + 4 = 4 \\ \Rightarrow x^{2} + y^{2} + 4 x + 4 y + 4 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n o f t h e c i r c l e i s x^{2} + y^{2} + 4 x + 4 y + 4 = 0 . \end{array}$

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