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New answer posted

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Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

If the lines $2 x - 3 y = 5$ and $3 x - 4 y = 7$ are the diameters of a circle with an area of 154 square units, then obtain the equation of the circle.

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} W e k n o w t h a t t h e intersection point o f t h e d i a m e t e r g i v e s t h e c e n t r e o f t h e c i r c l e . \\ G i v e n e q u a t i o n s o f d i a m e t e r s a r e \\ 2 x - 3 y = 5 \dots (i) \\ 3 x - 4 y = 7 \dots (i i) \\ F r o m e q n . (i) w e h a v e x = \frac{5 + 3 y}{2} \dots (i i i) \\ P u t t i n g t h e v a l u e o f x i n e q n . (i i) w e h a v e \\ 3 (\frac{5 + 3 y}{2}) - 4 y = 7 \\ \Rightarrow 15 + 9 y - 8 y = 14 \Rightarrow y = 14 - 15 \Rightarrow y = - 1 \\ N o w f r o m e q n . (i i i) w e h a v e x = \frac{5 + 3 (- 1)}{2} \Rightarrow x = \frac{5 - 3}{2} \Rightarrow x = 1 \\ S o, t h e c e n t r e o f t h e c i r c l e = (1, - 1) \\ G i v e n t h a t a r e a o f t h e c i r c l e = 154 \\ \Rightarrow π r^{2} = 154 \\ \Rightarrow \frac{22}{7} * r^{2} = 154 \Rightarrow r^{2} = 154 * \frac{7}{22} \\ \Rightarrow r^{2} = 7 * 7 \Rightarrow r = 7 \\ S o, t h e e q u a t i o n o f t h e c i r c l e i s \\ {(x - 1)}^{2} + {(y + 1)}^{2} = {(7)}^{2} \\ \Rightarrow x^{2} + 1 - 2 x + y^{2} + 1 + 2 y = 49 \\ \Rightarrow x^{2} + y^{2} - 2 x + 2 y = 47 \\ H e n c e, t h e r e q u i r e d e q u a t i o n o f t h e c i r c l e i s x^{2} + y^{2} - 2 x + 2 y = 47 . \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of the hyperbola with eccentricity $\sqrt[]{2}$ and foci at $(\pm 2, 0)$ .

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t e = \frac{3}{2} a n d f o c i = (\pm 2, 0) \\ W e k n o w t h a t f o c i = (\pm a e, 0) \\ ∴ a e = 2 \Rightarrow a * \frac{3}{2} = 2 \Rightarrow a = \frac{4}{3} \\ \Rightarrow a^{2} = \frac{16}{9} \\ W e k n o w t h a t b^{2} = a^{2} (e^{2} - 1) \\ \Rightarrow b^{2} = \frac{16}{9} (\frac{9}{4} - 1) = \frac{16}{9} * \frac{5}{4} = \frac{20}{9} \\ S o, t h e e q u a t i o n o f t h e h y p e r b o l a i s \\ \frac{x^{2}}{16 / 9} - \frac{y^{2}}{20 / 9} = 1 \\ \Rightarrow \frac{9 x^{2}}{16} - \frac{9 y^{2}}{20} = 1 \Rightarrow \frac{x^{2}}{4} - \frac{y^{2}}{5} = \frac{4}{9} \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s \frac{x^{2}}{4} - \frac{y^{2}}{5} = \frac{4}{9} . \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the eccentricity of the hyperbola $9 y^{2} - 4 x^{2} = 36$ .

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n e q u a t i o n i s 9 y^{2} - 4 x^{2} = 36 \\ \Rightarrow \frac{y^{2}}{4} - \frac{x^{2}}{9} = 36 \\ C l e a r l y i t i s v e r t i c a l h y p e r b o l a . \\ W h e r e a = 3 a n d b = 2 \\ N o w, b^{2} = a^{2} (e^{2} - 1) \\ \Rightarrow 4 = 9 (e^{2} - 1) \\ \Rightarrow e^{2} - 1 = \frac{4}{9} \Rightarrow e^{2} = 1 + \frac{4}{9} = \frac{13}{9} \\ ∴ e = \frac{\sqrt[]{13}}{3} \\ H e n c e, t h e r e q u i r e d v a l u e o f e i s \frac{\sqrt[]{13}}{3} . \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

If the distance between the foci of a hyperbola is 16 and its eccentricity $\sqrt[]{2}$ is, then obtain the equation of the hyperbola.

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} E q u a t i o n o f h y p e r b o l a i s \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \\ Distance b e t w e e n t h e f o c i = 2 a e \\ 2 a e = 16 \Rightarrow a e = 8 \\ \Rightarrow a * \sqrt[]{2} = 8 \Rightarrow a = \frac{8}{\sqrt[]{2}} = 4 \sqrt[]{2} [? e = \sqrt[]{2}] \\ N o w, b^{2} = a^{2} (e^{2} - 1) [F o r h y p e r b o l a] \\ \Rightarrow b^{2} = {(4 \sqrt[]{2})}^{2} (2 - 1) \\ \Rightarrow b^{2} = 32 \\ a = 4 \sqrt[]{2} \Rightarrow a^{2} = 32 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s \frac{x^{2}}{32} - \frac{y^{2}}{32} = 1 \Rightarrow x^{2} - y^{2} = 32 . \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

If the line $y = m x + 1$ is tangent to the parabola $y^{2} = 4 x$ , then find the value of $m$ .
[Hint: Solving the equation of the line and parabola, obtain a quadratic equation and apply the tangency condition to find $m$ ].

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t y^{2} = 4 x \dots (i) \\ a n d y = m x + 1 \dots (i i) \\ F r o m e q n . (i) a n d (i i) w e g e t \\ {(m x + 1)}^{2} = 4 x \\ \Rightarrow m^{2} x^{2} + 1 + 2 m x - 4 x = 0 \\ \Rightarrow m^{2} x^{2} + (2 m - 4) x + 1 = 0 \\ A p p l y i n g c o n d i t i o n o f tangency w e h a v e \\ {(2 m - 4)}^{2} - 4 m^{2} * 1 = 0 \\ \Rightarrow 4 m^{2} + 16 - 16 m - 4 m^{2} = 0 \\ \Rightarrow - 16 m = - 16 \Rightarrow m = 1 \\ H e n c e, t h e r e q u i r e d v a l u e o f m i s 1 . \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

If the points (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then find the equation of the parabola.

0 Follower 7 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t : \\ V e r t e x = (0, 4) a n d F o c u s = (0, 2) \\ L e t P (x, y) b e a n y point o n t h e p a r a b o l a . P B i s p e r p e n d i c u l a r t o t h e d i r e c t r i x . \\ A c c o r d i n g t o t h e d e f i n i t i o n o f p a r a b o l a, w e h a v e \\ P F = P B \\ \Rightarrow \sqrt[]{{(x - 0)}^{2} + {(y - 2)}^{2}} = | \frac{0 + y - 6}{\sqrt[]{0 + 1}} | [? E q u a t i o n o f d i r e c t r i x i s y = 6] \\ \Rightarrow \sqrt[]{x^{2} + {(y - 2)}^{2}} = (y - 6) \\ S q u a r i n g b o t h s i d e s, w e h a v e \\ x^{2} + y^{2} + 4 - 4 y = y^{2} + 36 - 12 y \\ \Rightarrow x^{2} - 4 y + 12 y - 32 = 0 \\ \Rightarrow x^{2} + 8 y - 32 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} + 8 y = 32 . \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the length of the line segment joining the vertex of the parabola $y^{2} = 4 a x$ and a point on the parabola where the line segment makes an angle $θ$ to the x-axis.

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n p a r a b o l a i s y^{2} = 4 a x \\ L e t P (a t^{2}, 2 a t) b e a n y o n t h e p a r a b o l a . \\ I n Δ P O A, \\ t a n θ = \frac{2 a t}{a t^{2}} = \frac{2}{t} \\ \Rightarrow t = \frac{2}{t a n θ} \\ \Rightarrow t = 2 c o t θ \dots (i) \\ O P = \sqrt[]{{(a t^{2} - 0)}^{2} + {(2 a t - 0)}^{2}} \\ = \sqrt[]{a^{2} t^{4} + 4 a^{2} t^{2}} \\ = a t \sqrt[]{t^{2} + 4} \\ = a * 2 c o t θ \sqrt[]{4^{} + 4} [? t = 2 c o t θ] \\ = 2 a c o t θ . 2 \sqrt[]{c o t^{2} θ + 1} = 4 a c o t θ c o s e c θ \\ = 4 a . \frac{c o s θ}{s i n θ} . \frac{1}{s i n θ} = \frac{4 a c o s θ}{s i n^{2} θ} \\ H e n c e, t h e r e q u i r e d l e n g t h = \frac{4 a c o s θ}{s i n^{2} θ} . \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the coordinates of a point on the parabola $y^{2} = 8 x$ whose focal distance is 4.

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n p a r a b o l a i s y^{2} = 8 x \dots (i) \\ C o m p a r i n g w i t h t h e e q u a t i o n o f p a r a b o l a y^{2} = 4 a x \\ 4 a = 8 \Rightarrow a = 2 \\ N o w f o c a lDistance = | x + a | \\ \Rightarrow | x + a | = 4 \\ \Rightarrow (x + a) = \pm 4 \Rightarrow x + 2 = \pm 4 \\ \Rightarrow x = 4 - 2 = 2 a n d x = - 6 \\ B u t x \neq - 6 ∴ x = 2 \\ P u t x = 2 i n e q u a t i o n (i) w e g e t \\ y^{2} = 8 * 2 = 16 \\ ∴ y = \pm 4 \\ S o, t h e c o o r d i n a t e s o f t h e points a r e (2, 4), (2, - 4) . \\ H e n c e, t h e r e q u i r e d c o o r d i n a t e s a r e (2, 4), (2, - 4) . \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

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

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

a year ago

Class 11th 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)

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

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