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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

An ellipse is described by using an endless string that passes over two pins. If the axes are 6 cm and 4 cm, the length of the string and the distance between the pins are ______.

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alok kumar singh

Contributor-Level 10

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t t h e e q u a t i o n o f a n e l l i p s e i s \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \\ H e r e, 2 a = 6 \Rightarrow a = 3 \\ a n d 2 b = 4 \Rightarrow b = 2 \\ W e k n o w t h a t c^{2} = a^{2} - b^{2} \\ = {(3)}^{2} - {(2)}^{2} = 9 - 4 = 5 \\ ∴ c = \sqrt[]{5}, w e h a v e e = \frac{c}{a} \Rightarrow e = \frac{\sqrt[]{5}}{3} \\ L e n g t h o f s t r i n g = 2 a + 2 a e = 2 a (1 + e) \\ = 6 (1 + \frac{\sqrt[]{5}}{3}) = \frac{6 (3 + \sqrt[]{5})}{3} = 6 + 2 \sqrt[]{5} \\ Distance b e t w e e n t h e p i n s = C C' = 2 a e = 2 * 3 * \frac{\sqrt[]{5}}{3} = 2 \sqrt[]{5} \\ H e n c e, t h e v a l u e o f f i l l e r a r e 6 + 2 \sqrt[]{5} c m a n d 2 \sqrt[]{5} c m . \end{array}$

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3 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The equation of the circle circumscribing the triangle whose sides are the lines $y = x + 2$ , $3 y = 4 x$ , and $2 y = 3 x$ is ______.

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alok kumar singh

Contributor-Level 10

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t A B r e p r e s e n t s 2 y = 3 x \dots (i) \\ B C r e p r e s e n t s 3 y = 4 x \dots (i i) \\ a n d A C r e p r e s e n t s y = x + 2 \dots (i i i) \\ F r o m e q n . (i) a n d (i i) \\ 2 y = 3 x \Rightarrow y = \frac{3 x}{2} \\ P u t t i n g t h e v a l u e o f y i n e q n . (i i) w e g e t \\ 3 (\frac{3 x}{2}) = 4 x \Rightarrow 9 x = 8 x \\ \Rightarrow x = 0 a n d y = 0 \\ ∴ C o o r d i n a t e s o f B (0, 0) \\ F r o m e q n . (i) a n d (i i i) w e g e t \\ y = x + 2 \\ P u t t i n g y = x + 2 i n e q n . (i) w e g e t \\ 2 (x + 2) = 3 x \\ \Rightarrow 2 x + 4 = 3 x \Rightarrow x = 4 a n d y = 6 \\ ∴ C o o r d i n a t e s o f A (4, 6) \\ S o l v i n g e q n . (i i) a n d (i i i) w e g e t \\ y = x + 2 \\ P u t t i n g t h e v a l u e o f y i n e q n . (i i) w e g e t \\ 3 (x + 2) = 4 x \Rightarrow 3 x + 6 = 4 x \Rightarrow x = 6 a n d y = 8 \\ ∴ C o o r d i n a t e s o f C (6, 8) \\ I t i m p l i e s t h a t t h e c i r c l e i s t h r o u g h (0, 0), (4, 6) a n d (6, 8) . \\ W e k n o w t h a t t h e g e n e r a l e q u a t i o n o f t h e c i r c l e i s \\ x^{2} + y^{2} + 2 g x + 2 f y + c = 0 \dots (i v) \\ Since t h e points (0, 0), (4, 6) a n d (6, 8) l i e o n t h e c i r c l e t h e n \\ 0 + 0 + 0 + 0 + c = 0 \Rightarrow c = 0 \\ 16 + 36 + 8 g + 12 f + c = 0 \Rightarrow 8 g + 12 f + 0 = - 52 \\ \Rightarrow &th \end{array}$

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New answer posted

3 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The equation of the circle having its center at (3,-4) and touching the line $5 x + 12 y - 12 = 0$ is ______.

[Hint: To determine the radius, find the perpendicular distance from the center of the circle to the line.]

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n o f t h e l i n e i s 5 x + 12 y - 12 = 0 a n d t h e c e n t r e i s (3, - 4) \\ C P = r a d i u s o f t h e c i r c l e \\ | \frac{5 * 3 + 12 * (- 4) - 12}{\sqrt[]{{(5)}^{2} + {(12)}^{2}}} | = r \\ \Rightarrow | \frac{15 - 48 - 12}{13} | = r \\ \Rightarrow | \frac{- 45}{13} | = r \Rightarrow r^{2} = \frac{2025}{169} \\ S o, t h e e q u a t i o n o f t h e c i r c l e i s {(x - 3)}^{2} + {(y + 4)}^{2} = {(\frac{45}{13})}^{2} \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s {(x - 3)}^{2} + {(y + 4)}^{2} = {(\frac{45}{13})}^{2} . \end{array}$

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New answer posted

3 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The locus of the point of intersection of lines $\sqrt[]{3} x - y - 4 \sqrt[]{3} k = 0$ and $\sqrt[]{3} k x + k y - 4 \sqrt[]{3} = 0$ for different values of $k$ is a hyperbola whose eccentricity is 2.

[Hint: Eliminate $k$ between the given equations.]

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} T h e g i v e n e q u a t i o n s a r e \\ \sqrt[]{3} x - y - 4 \sqrt[]{3} k = 0 \dots (i) \\ a n d \sqrt[]{3} k x + k y - 4 \sqrt[]{3} = 0 \dots (i i) \\ F r o m e q n . (i) w e g e t \\ 4 \sqrt[]{3} k = \sqrt[]{3} x - y \\ ∴ k = \frac{\sqrt[]{3} x - y}{4 \sqrt[]{3}} \\ P u t t i n g t h e v a l u e o f k i n e q n . (i i), w e g e t \\ \sqrt[]{3} [\frac{\sqrt[]{3} x - y}{4 \sqrt[]{3}}] x + [\frac{\sqrt[]{3} x - y}{4 \sqrt[]{3}}] y - 4 \sqrt[]{3} = 0 \\ \Rightarrow (\frac{\sqrt[]{3} x - y}{4}) x + (\frac{\sqrt[]{3} x - y}{4 \sqrt[]{3}}) y - 4 \sqrt[]{3} = 0 \\ \Rightarrow \frac{(3 x - \sqrt[]{3} y) x + (\sqrt[]{3} x - y) y - 48}{4 \sqrt[]{3}} = 0 \\ \Rightarrow 3 x^{2} - \sqrt[]{3} x y + \sqrt[]{3} x y - y^{2} - 48 = 0 \\ \Rightarrow 3 x^{2} - y^{2} = 48 \\ \Rightarrow \frac{x^{2}}{16} - \frac{y^{2}}{48} = 1 (w h i c h i s a h y p e r b o l a) \\ H e r e a^{2} = 16, b^{2} = 48 \\ W e k n o w t h a t b^{2} = a^{2} (e^{2} - 1) \\ \Rightarrow 48 = 16 (e^{2} - 1) \\ \Rightarrow 3 = e^{2} - 1 \Rightarrow e^{2} = 4 \Rightarrow e = 2 \\ H e n c e, t h e g i v e n s t a t e m e n t i s T r u e . \end{array}$

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New answer posted

3 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The line $2 x + 3 y = 12$ touches the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ at the point (3, 2). True or False?

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} I f l i n e 2 x + 3 y = 12 t o u c h e s t h e e l l i p s e \frac{x^{2}}{9} + \frac{y^{2}}{4} = 2, \\ t h e n t h e point (3, 2) s a t i s f i e s b o t h t h e l i n e a n d e l l i p s e . \\ ∴ F o r l i n e 2 x + 3 y = 12 \\ 2 (3) + 3 (2) = 12 \Rightarrow 6 + 6 = 12 \Rightarrow 12 = 12 T r u e \\ F o r e l l i p s e \frac{x^{2}}{9} + \frac{y^{2}}{4} = 2 \\ \Rightarrow \frac{{(3)}^{2}}{9} + \frac{{(2)}^{2}}{4} = 2 \Rightarrow \frac{9}{9} + \frac{4}{4} = 2 \Rightarrow 1 + 1 = 2 \Rightarrow 2 = 2 T r u e \\ H e n c e, t h e g i v e n s t a t e m e n t i s T r u e . \end{array}$

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New answer posted

3 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

If $P$ is a point on the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{25} = 1$ , whose foci are $S$ and $S^{'}$ , then $P S + P S^{'} = 8$

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t P (x_{1}, y_{1}) b e apoint on o n t h e e l l i p s e . \\ f o c i = (\pm a e, 0) \\ H e r e, a^{2} = 25 \Rightarrow a = 5 \\ b^{2} = 16 \Rightarrow b = 4 \\ b^{2} = a^{2} (1 - e^{2}) \\ 16 = 25 (1 - e^{2}) \\ \Rightarrow \frac{16}{25} = 1 - e^{2} \Rightarrow e^{2} = 1 - \frac{16}{25} \Rightarrow e^{2} = \frac{9}{25} \\ ∴ e = \frac{3}{5} \\ ∴ a e = 5 * \frac{3}{5} = 3 \\ S o, t h e f o c i a r e S (3, 0) a n d S' (- 3, 0) . \\ Since P S + P S' = 2 a = 2 * 5 = 10 . \\ H e n c e, t h e g i v e n s t a t e m e n t i s F a l s e . \end{array}$

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New answer posted

3 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The line $l x + m y + n = 0$ will touch the parabola $y^{2} = 4 a x$ if $l n = a m^{2}$ . True or False?

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n e q u a t i o n o f p a r a b o l a i s y^{2} = 4 a x \dots (i) \\ a n d t h e e q u a t i o n o f l i n e i s l x + m y + n = 0 \dots (i i) \\ F r o m e q n . (i i), w e h a v e \\ y = \frac{- l x - n}{m} \\ P u t t i n g t h e v a l u e o f y i n e q n . (i) w e g e t \\ {(\frac{- l x - n}{m})}^{2} = 4 a x \\ \Rightarrow l^{2} x^{2} + n^{2} + 2 l n x - 4 a m^{2} x = 0 \\ \Rightarrow l^{2} x^{2} + (2 l n - 4 a m^{2}) x + n^{2} = 0 \\ I f t h e l i n e i s tangent t h e t o t h e c i r c l e, t h e n \\ b^{2} - 4 a c = 0 \\ \Rightarrow {(2 l n - 4 a m^{2})}^{2} - 4 l^{2} n^{2} = 0 \\ \Rightarrow 4 l^{2} n^{2} + 16 a^{2} m^{4} - 16 l n m^{2} a - 4 l^{2} n^{2} = 0 \\ \Rightarrow 16 a^{2} m^{4} - 16 l n m^{2} a = 0 \\ \Rightarrow 16 a m^{2} (a m^{2} - l n) = 0 \\ \Rightarrow a m^{2} (a m^{2} - l n) = 0 \\ \Rightarrow a m^{2} \neq 0 ∴ a m^{2} - l n = 0 \\ \Rightarrow l n = a m^{2} \\ H e n c e, t h e g i v e n s t a t e m e n t i s T r u e . \end{array}$

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New answer posted

3 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The point (1, 2) lies inside the circle $x^{2} + y^{2} - 2 x + 6 y + 1 = 0$ . True or False?

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n e q u a t i o n o f t h e c i r c l e i s x^{2} + y^{2} - 2 x + 6 y + 1 = 0 \\ H e r e, 2 g = - 2 \Rightarrow g = - 1 \\ 2 f = 6 \Rightarrow f = 3 \\ ∴ C e n t r e = (- g, - f) = (1, - 3) \\ a n d r a d i u s r = \sqrt[]{g^{2} + f^{2} - c} = \sqrt[]{1 + 9 - 1} = 3 \\ ∴ Distance b e t w e e n t h e point (1, 2) a n d t h e c e n t r e (1, - 3) \\ = \sqrt[]{{(1 - 1)}^{2} + {(2 + 3)}^{2}} = 5 \\ H e r e, 5 > 3, s o t h e point l i e s o u t s i d e t h e c i r c l e . \\ H e n c e, t h e g i v e n s t a t e m e n t i s F a l s e . \end{array}$

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New answer posted

3 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

If the line $l x + m y = 1$ is a tangent to the circle $x^{2} + y^{2} = a^{2}$ , then the point (l, m) lies on a circle. True or False?

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is a true and false Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n o f t h e c i r c l e i s x^{2} + y^{2} = a^{2} \\ a n d t h e tangent i s l x + m y = 1 \\ H e r e c e n t r e i s (0, 0) a n d r a d i u s = a \\ I f (l, m) l i e s o n t h e c i r c l e \\ ∴ \sqrt[]{{(l - 0)}^{2} + {(m - 0)}^{2}} = a \\ \Rightarrow \sqrt[]{l^{2} + m^{2}} = a \Rightarrow l^{2} + m^{2} = a^{2} (w h i c h i s a c i r c l e) \\ S o, t h e (l, m) l i e s o n t h e c i r c l e . \\ H e n c e, t h e g i v e n s t a t e m e n t i s T r u e . \end{array}$

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New answer posted

3 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The shortest distance from the point (2, -7) to the circle $x^{2} + y^{2} - 14 x - 10 y - 151 = 0$ is equal to 5.

[Hint: The shortest distance is equal to the difference of the radius and the distance between the center and the given point.]

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is a true and false Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n o f t h e c i r c l e i s x^{2} + y^{2} - 14 x - 10 y - 151 = 0 \\ S h o r t e s t distance =distance b e t w e e n t h e (2, - 7) \\ a n d t h e c e n t r e - r a d i u s o f t h e c i r c l e \\ C e n t r e o f t h e g i v e n c i r c l e i s \\ 2 g = - 14 \Rightarrow g = - 7 \\ 2 f = - 10 \Rightarrow f = - 5 \\ ∴ C e n t r e = (- g, - f) = (7, 5) \\ a n d r = \sqrt[]{{(- 7)}^{2} + {(- 5)}^{2} + 151} = \sqrt[]{49 + 25 + 151} \\ = \sqrt[]{225} = 15 \\ ∴ S h o r t e s t = \sqrt[]{{(7 - 2)}^{2} + {(5 + 7)}^{2}} - 15 \\ = \sqrt[]{25 + 144} - 15 = 13 - 15 = | - 2 | = 2 \\ H e n c e, t h e g i v e n s t a t e m e n t i s F a l s e . \end{array}$

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