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Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

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Maths NCERT Exemplar Solutions Class 11th Chapter Eleven Conic Sections

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.]

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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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Maths NCERT Exemplar Solutions Class 11th Chapter Eleven Exemplar Solution

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?

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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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Maths NCERT Exemplar Solutions Class 11th Chapter Eleven Conic Sections

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

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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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Maths NCERT Exemplar Solutions Class 11th Chapter Eleven Conic Sections

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?

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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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Maths NCERT Exemplar Solutions Class 11th Chapter Eleven Conic Sections

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

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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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Maths NCERT Exemplar Solutions Class 11th Chapter Eleven Exemplar Solution

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?

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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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Maths NCERT Exemplar Solutions Class 11th Chapter Eleven Conic Sections

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.]

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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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Maths NCERT Exemplar Solutions Class 11th Chapter Eleven Conic Sections

The line $x + 3 y = 0$ is a diameter of the circle $x^{2} + y^{2} + 6 x + 2 y = 0$ . True or False?

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n e q u a t i o n o f t h e c i r c l e i s \\ x^{2} + y^{2} + 6 x + 2 y = 0 \\ C e n t r e i s (- 3, - 1) \\ I f x + 3 y = 0 i s t h e e q u a t i o n o f d i a m e t e r, t h e n t h e c e n t r e (- 3, - 1) w i l l l i e o n x + 3 y = 0 \\ - 3 + 3 (- 1) = 0 \\ \Rightarrow - 6 \neq 0 \\ S o, x + 3 y = 0 i s n o t t h e d i a m e t e r o f t h e c i r c l e . \\ H e n c e, t h e g i v e n s t a t e m e n t i s F a l s e . \end{array}$

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Maths NCERT Exemplar Solutions Class 11th Chapter Eleven Conic Sections

Find the equation of the hyperbola with:

Vertices $(\pm 5, 0)$ , foci $(\pm 7, 0)$ .

Vertices $(0, \pm 7)$ , eccentricity $e = \sqrt[]{2}$ .

Foci $(0, \pm 10)$ , passing through (2, 3).

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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} (i) G i v e n t h a t v e r t i c e s (\pm 5, 0), f o c i (\pm 7, 0) \\ V e r t e x o f h y p e r b o l a = (\pm a, 0) a n d f o c i (\pm a e, 0) \\ ∴ a = 5 a n d a e = 7 \Rightarrow 5 * e = 7 \Rightarrow e = \frac{7}{5} \\ N o w b^{2} = a^{2} (e^{2} - 1) \\ \Rightarrow b^{2} = 25 (\frac{49}{25} - 1) \Rightarrow b^{2} = 25 * \frac{24}{25} \Rightarrow b^{2} = 24 \\ 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}}{25} - \frac{y^{2}}{24} = 1 \\ (i i) G i v e n t h a t v e r t i c e s (0, \pm 7), e = \frac{4}{3} \\ C l e a r l y, t h e h y p e r b o l a i s v e r t i c a l . \\ ∴ a = 5 a n d e = \frac{4}{3} \\ W e k n o w t h a t b^{2} = a^{2} (e^{2} - 1) \\ \Rightarrow b^{2} = 49 (\frac{16}{9} - 1) \Rightarrow b^{2} = 49 * \frac{7}{9} \Rightarrow b^{2} = \frac{343}{9} \\ 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{y^{2}}{49} - \frac{9 x^{2}}{343} = 1 \Rightarrow 9 x^{2} - 7 y^{2} + 343 = 0 \\ (i i i) G i v e n t h a t : f o c i (0, \pm \sqrt[]{10}) \\ ∴ a e = \sqrt[]{10} \Rightarrow a^{2} e^{2} = 10 \\ W e k n o w t h a t b^{2} = a^{2} (e^{2} - 1) \\ \Rightarrow b^{2} = a^{2} e^{2} - a^{2} \Rightarrow b^{2} = 10 - a^{2} \\ 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{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \Rightarrow \frac{y^{2}}{a^{2}} - \frac{x^{2}}{10 - a^{2}} = 1 \\ I f i t p a s s e s t h r o u g h t h e (2, 3) t h e n \\ \Rightarrow \frac{9}{a^{2}} - \frac{4}{10 - a^{2}} = 1 \Rightarrow \frac{90 - 9 a^{2} - 4 a^{2}}{a^{2} (10 - a^{2})} = 1 \\ \Rightarrow 90 - 13 a^{2} = 10 a^{2} - a^{4} \Rightarrow a^{4} - 23 a^{2} + 90 = 0 \\ \Rightarrow a^{4} - 18 a^{2} - 5 a^{2} + 90 = 0 \\ \Rightarrow a^{2} (a^{2} - 18) - 5 (a^{2} - 18) = 0 \\ \Rightarrow (a^{2} - 18) (a^{2} - 5) = 0 \\ \Rightarrow a^{2} = 18, a^{2} = 5 \\ ∴ b^{2} = 10 - 18 = - 8 a n d b^{2} = 10 - 5 = 5 \\ ∴ b \neq - 8 a n d b^{2} = 5 \\ 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{y^{2}}{5} - \frac{x^{2}}{5} = 1 o r y^{2} - x^{2} = 5 . \end{array}$

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Maths NCERT Exemplar Solutions Class 11th Chapter Eleven Conic Sections

Show that the set of all points such that the difference of their distances from (4, 0) and (-4, 0) is always equal to 2 represents a hyperbola.

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t (x, y) b e a n ypoint \\ A c c o r d i n g t o t h e q u e s t i o n, w e h a v e \\ \sqrt[]{{(x - 4)}^{2} + {(y - 0)}^{2}} - \sqrt[]{{(x + 4)}^{2} + {(y - 0)}^{2}} = 2 \\ \Rightarrow \sqrt[]{x^{2} + 16 - 8 x + y^{2}} - \sqrt[]{x^{2} + 16 + 8 x + y^{2}} = 2 \\ P u t t i n g x^{2} + y^{2} + 16 = z \dots (i) \\ \Rightarrow \sqrt[]{z - 8 x} - \sqrt[]{z + 8 x} = 2 \\ S q u a r i n g b o t h s i d e s, w e h a v e \\ \Rightarrow z - 8 x + z + 8 x - 2 \sqrt[]{(z - 8 x) (z + 8 x)} = 4 \\ \Rightarrow 2 z - 2 \sqrt[]{z^{2} - 64 x^{2}} = 4 \\ \Rightarrow z - \sqrt[]{z^{2} - 64 x^{2}} = 2 \\ \Rightarrow (z - 2) = \sqrt[]{z^{2} - 64 x^{2}} \\ A g a i n s q u a r i n g b o t h s i d e s, w e g e t \\ \Rightarrow z^{2} + 4 - 4 z = z^{2} - 64 x^{2} \\ \Rightarrow 4 - 4 z + 64 x^{2} = 0 \\ P u t t i n g t h e v a l u e o f z, w e h a v e \\ \Rightarrow 4 - 4 (x^{2} + y^{2} + 16) + 64 x^{2} = 0 \\ \Rightarrow 4 - 4 x^{2} - 4 y^{2} - 64 + 64 x^{2} = 0 \\ \Rightarrow 60 x^{2} - 4 y^{2} - 60 = 0 \\ \Rightarrow 60 x^{2} - 4 y^{2} = 60 \\ \Rightarrow \frac{60 x^{2}}{60} - \frac{4 y^{2}}{60} = 1 \\ \Rightarrow \frac{x^{2}}{1} - \frac{y^{2}}{15} = 1 \\ W h i c h r e p r e s e n t a h p e r b o l a . H e n c e p r o v e d . \end{array}$

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