Conic Sections

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

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

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 Short Answer Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year 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)

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year 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.

9 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year 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.

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year 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.

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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•a year 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.]

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year 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.]

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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•a year 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 $3 x + y = 14$ , $2 x + 5 y = 18$ .

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year 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.

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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•a year 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.]

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 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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