Conic Sections

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The equation of the parabola having focus at (â€“1, â€“2) and the directrix $x - 2 y + 3 = 0$ is ______.

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} L e t (x_{1}, y_{1}) b e a n y point o n t h e p a r a b o l a . \\ A c c o r d i n g t o t h e d e f i n i t i o n o f t h e p a r a b o l a \\ \sqrt[]{{(x_{1} + 1)}^{2} + {(y_{1} + 2)}^{2}} = | \frac{x_{1} - 2 y_{1} + 3}{\sqrt[]{{(1)}^{2} + {(- 2)}^{2}}} | \\ S q u a r i n g b o t h s i d e s, w e g e t \\ x_{1}^{2} + 1 + 2 x_{1} + y_{1}^{2} + 4 + 4 y_{1} = \frac{x_{1}^{2} + 4 y_{1}^{2} + 9 - 4 x_{1} y_{1} - 12 y_{1} + 6 x_{1}}{5} \\ \Rightarrow x_{1}^{2} + y_{1}^{2} + 2 x_{1} + 4 y_{1} + 5 = \frac{x_{1}^{2} + 4 y_{1}^{2} - 4 x_{1} y_{1} - 12 y_{1} + 6 x_{1} + 9}{5} \\ \Rightarrow 5 x_{1}^{2} + 5 y_{1}^{2} + 10 x_{1} + 20 y_{1} + 25 = x_{1}^{2} + 4 y_{1}^{2} - 4 x_{1} y_{1} - 12 y_{1} + 6 x_{1} + 9 \\ \Rightarrow 4 x_{1}^{2} + y_{1}^{2} + 4 x_{1} + 32 y_{1} + 4 x_{1} y_{1} + 16 = 0 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 4 x^{2} + 4 x y + y^{2} + 4 x + 32 y + 16 = 0 . \end{array}$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The equation of the ellipse having foci (0, 1), (0, â€“1), and a minor axis of length 1 is ______.

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year ago

Conic Sections 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 ______.

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 Fill in the Blanks Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year ago

Conic Sections 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 ______.

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 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•a year ago

Conic Sections 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.]

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year ago

Conic Sections 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.]

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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•a year ago

Conic Sections 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?

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 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•a year ago

Conic Sections 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 $P S + P S^{'} = 8$

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 True and False Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year ago

Conic Sections 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?

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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•a year ago

Conic Sections 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?

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 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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