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New answer posted•a year ago

Exemplar Solution 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

Exemplar Solution 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

Exemplar Solution 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

Exemplar Solution 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

Exemplar Solution 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

Exemplar Solution 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

Exemplar Solution 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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New answer posted•a year ago

Exemplar Solution 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.]

4 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:

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen

The sum of probabilities of two students getting distinction in their final examinations is 1.2.

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} Since, t h e t w o g i v e n e v e n t s a r e n o t r e l a t e d t o t h e s a m e S a m p l e s p a c e . \\ ∴ T h e s u m o f p r o b a b i l i t i e s o f t w o s t u d e n t s g e t t i n g d i s t i n c t i o n i n t h e i r f i n a l \\ m a y b e 1.2 \\ H e n c e, t h e g i v e n s t a t e m e n t i s' T r u e' \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen

The probability of the occurrence of event A is 0.7 and that of the occurrence of event B is 0.3, but the probability of the occurrence of both is 0.4.

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} H e r e, P (A) = 0.7, P (B) = 0.3 \\ ∴ P (A \cap B) = P (A) \times P (B) \\ = 0.7 \times 0.3 = 0.21 \\ B u t t h e g i v e n p r o b a b i l i t y i s 0.4 \\ H e n c e, t h e g i v e n s t a t e m e n t i s' F a l s e' \end{array}$

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