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Maths 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

a year ago

Maths 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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New answer posted

a year ago

Maths 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

a year ago

Maths 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

a year ago

Maths 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.]

0 Follower 2 Views

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

a year ago

Maths 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.]

0 Follower 2 Views

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

a year ago

Maths 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 $3 x + y = 14$ , $2 x + 5 y = 18$ .

0 Follower 2 Views

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

a year ago

Maths 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.

0 Follower 2 Views

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

a year ago

Maths 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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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

If the lines $3 x - 4 y + 4 = 0$ and $6 x - 8 y - 7 = 0$ are tangents to a circle, then find the radius of the circle.
[Hint: Distance between given parallel lines gives the diameter of the circle.]

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

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

Sol:

Sol:

$G i v e n e q u a t i o n a r e 3 x - 4 y + 4 = 0 a n d 6 x - 8 y - 7 = 0 \Rightarrow 3 x - 4 y - \frac{7}{2} = 0$
$\begin{array}{l} Since \frac{3}{6} = \frac{- 4}{- 8} = \frac{1}{2} t h e n t h e l i n e s a r e p a r a l l e l . \\ S o, t h e b e t w e e n t h e p a r a l l e l l i n e s = | \frac{c_{1} - c_{2}}{\sqrt[]{a^{2} + b^{2}}} | = | \frac{4 + \frac{7}{2}}{\sqrt[]{{(3)}^{2} + {(- 4)}^{2}}} | \\ = | \frac{\frac{15}{2}}{5} | = \frac{3}{2} \\ D i a m e t e r = \frac{3}{2} \\ ∴ R a d i u s = \frac{3}{4} \\ H e n c e, t h e r e q u i r e d r a d i u s = \frac{3}{4} . \end{array}$

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