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2 months ago

Conic Sections 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?

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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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New answer posted

2 months ago

Conic Sections 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.]

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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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2 months ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

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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New answer posted

2 months ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

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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New answer posted

2 months ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

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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New answer posted

2 months ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of the set of all points whose distance from (0, 4) equals their distance from the line $y = 9$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t (x, y) b e a n y \\ A c c o r d i n g t o t h e q u e s t i o n, w e h a v e \\ \sqrt[]{{(x - 0)}^{2} + {(y - 4)}^{2}} = \frac{2}{3} | \frac{y - 9}{1} | \\ S q u a r i n g b o t h s i d e s, w e h a v e \\ \Rightarrow x^{2} + {(y - 4)}^{2} = \frac{4}{9} (y^{2} + 81 - 18 y) \\ \Rightarrow 9 x^{2} + 9 {(y - 4)}^{2} = 4 y^{2} + 324 - 72 y \\ \Rightarrow 9 x^{2} + 9 y^{2} + 144 - 72 y = 4 y^{2} + 324 - 72 y \\ \Rightarrow 9 x^{2} + 5 y^{2} + 144 - 324 = 0 \\ \Rightarrow 9 x^{2} + 5 y^{2} - 180 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s 9 x^{2} + 5 y^{2} - 180 = 0 . \end{array}$

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New answer posted

2 months ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of the set of all points the sum of whose distances from the points (3, 0) and (9, 0) is 12.

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t (x, y) b e a n y \\ G i v e n a r e (3, 0) a n d (9, 0) \\ A c c o r d i n g t o t h e q u e s t i o n, w e h a v e \\ \sqrt[]{{(x - 3)}^{2} + {(y - 0)}^{2}} + \sqrt[]{{(x - 9)}^{2} + {(y - 0)}^{2}} = 12 \\ \Rightarrow \sqrt[]{x^{2} + 9 - 6 x + y^{2}} + \sqrt[]{x^{2} + 81 - 18 x + y^{2}} = 12 \\ P u t t i n g x^{2} + 9 - 6 x + y^{2} = k \\ \Rightarrow \sqrt[]{k} + \sqrt[]{72 - 12 x + k} = 12 \\ \Rightarrow \sqrt[]{72 - 12 x + k} = 12 - \sqrt[]{k} \\ S q u a r i n g b o t h s i d e s, w e h a v e \\ \Rightarrow 72 - 12 x + k = 144 + k - 24 \sqrt[]{k} \\ \Rightarrow 24 \sqrt[]{k} = 144 - 72 + 12 x \\ \Rightarrow 24 \sqrt[]{k} = 72 + 12 x \\ \Rightarrow 2 \sqrt[]{k} = 6 + x \\ A g a i n s q u a r i n g b o t h s i d e s, w e g e t \\ \Rightarrow 4 k = 36 + x^{2} + 12 x \\ P u t t i n g t h e v a l u e o f k, w e h a v e \\ 4 (x^{2} + 9 - 6 x + y^{2}) = 36 + x^{2} + 12 x \\ \Rightarrow 4 x^{2} + 36 - 24 x + 4 y^{2} = 36 + x^{2} + 12 x \\ \Rightarrow 3 x^{2} + 4 y^{2} - 36 x = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s 3 x^{2} + 4 y^{2} - 36 x = 0 . \end{array}$

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New answer posted

2 months ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of each of the following parabolas:

Directrix $x = 0$ , focus at (6, 0).

Vertex at (0, 4), focus at (0, 2).

Focus at (-1,-2), directrix $x - 2 y + 3 = 0$ .

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} (i) G i v e n t h a t d i r e c t r i x = 0 a n d f o c u s (6, 0) \\ ∴ T h e e q u a t i o n o f t h e p a r a b o l a i s \\ {(x - 6)}^{2} + y^{2} = x^{2} \\ \Rightarrow x^{2} + 36 - 12 x + y^{2} = x^{2} \\ \Rightarrow y^{2} - 12 x + 36 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s y^{2} - 12 x + 36 = 0 \\ (i i) G i v e n t h a t v e r t e x a t (0, 4) a n d f o c u s a t (0, 2) . \\ S o, e q u a t i o n o f d i r e c t r i x i s y - 6 = 0 \\ A c c o r d i n g t o t h e d e f i n i t i o n o f t h e p a r a b o l a \\ P F = P M \\ \sqrt[]{{(x - 0)}^{2} + {(y - 2)}^{2}} = | y - 6 | \\ \Rightarrow \sqrt[]{x^{2} + y^{2} + 4 - 4 y} = | y - 6 | \\ S q u a r i n g b o t h s i d e s, w e g e t \\ x^{2} + y^{2} + 4 - 4 y = y^{2} + 36 - 12 y \\ \Rightarrow x^{2} + 4 - 4 y = 36 - 12 y \\ \Rightarrow x^{2} + 8 y - 32 = 0 \\ \Rightarrow x^{2} = 32 - 8 y \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} = 32 - 8 y \\ (i i i) G i v e n t h a t f o c u s a t (- 1, - 2) a n d d i r e c t r i x x - 2 y + 3 = 0 \\ L e t (x, y) b e a n y o n t h e p a r a b o l a . \\ A c c o r d i n g t o t h e d e f i n i t i o n o f t h e p a r a b o l a \\ P F = P M \\ \sqrt[]{{(x + 1)}^{2} + {(y + 2)}^{2}} = | \frac{x - 2 y + 3}{\sqrt[]{{(1)}^{2} + {(- 2)}^{2}}} | \\ \Rightarrow \sqrt[]{x^{2} + 1 + 2 x + y^{2} + 4 + 4 y} = | \frac{x - 2 y + 3}{\sqrt[]{5}} | \\ S q u a r i n g b o t h s i d e s, w e g e t \\ x^{2} + 1 + 2 x + y^{2} + 4 + 4 y = \frac{x^{2} + 4 y^{2} + 9 - 4 x y - 12 y + 6 x}{5} \\ \Rightarrow 5 x^{2} + 5 + 10 x + 5 y^{2} + 20 + 20 y = x^{2} + 4 y^{2} + 9 - 4 x y - 12 y + 6 x \\ \Rightarrow 4 x^{2} + y^{2} + 4 x y + 4 x + 32 y + 16 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s 4 x^{2} + y^{2} + 4 x y + 4 x + 32 y + 16 = 0 . \end{array}$

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New answer posted

2 months ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of a circle passing through the point (7, 3), having a radius of 3 units, and whose center lies on the line $y = x - 1$ .

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t t h e e q u a t i o n o f t h e c i r c l e b e \\ {(x - h)}^{2} + {(y - k)}^{2} = r^{2} \\ I f i t p a s s e s t h r o u g h (7, 3) t h e n \\ {(7 - h)}^{2} + {(3 - k)}^{2} = {(3)}^{2} [? r = 3] \\ \Rightarrow 49 + h^{2} - 14 h + 9 + k^{2} - 6 k = 9 \\ \Rightarrow h^{2} + k^{2} - 14 h - 6 k + 49 = 0 \dots (i) \\ I f c e n t r e (h, k) l i e s o n t h e l i n e y = x - 1 t h e n \\ k = h - 1 \dots (i i) \\ P u t t i n g t h e v a l u e o f k i n e q n . (i) w e g e t \\ h^{2} + {(h - 1)}^{2} - 14 h - 6 (h - 1) + 49 = 0 \\ \Rightarrow h^{2} + h^{2} + 1 - 2 h - 14 h - 6 h + 6 + 49 = 0 \\ \Rightarrow 2 h^{2} - 22 h + 56 = 0 \\ \Rightarrow 2 (h^{2} - 11 h + 28) = 0 \\ \Rightarrow h^{2} - 11 h + 28 = 0 \\ \Rightarrow h^{2} - 7 h - 4 h + 28 = 0 \\ \Rightarrow h (h - 7) - 4 (h - 7) = 0 \\ \Rightarrow (h - 7) (h - 4) = 0 \Rightarrow h = 7, 4 \\ F r o m e q n . (i i) w e g e t k = 4 - 1 = 3 a n d k = 7 - 1 = 6 . \\ S o, t h e c e n t r e s a r e (4, 3) a n d (7, 6) . \\ E q u a t i o n o f t h e c i r c l e i s \\ T a k i n g c e n t r e (4, 3) \\ {(x - 4)}^{2} + {(y - 3)}^{2} = 9 \\ \Rightarrow x^{2} + 16 - 8 x + y^{2} + 9 - 6 y = 9 \\ \Rightarrow x^{2} + y^{2} - 8 x - 6 y + 16 = 0 \\ T a k i n g c e n t r e (7, 6) \\ {(x - 7)}^{2} + {(y - 6)}^{2} = 9 \\ \Rightarrow x^{2} + 49 - 14 x + y^{2} + 36 - 12 y = 9 \\ \Rightarrow x^{2} + y^{2} - 14 x - 12 y + 76 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n s a r e \\ x^{2} + y^{2} - 8 x - 6 y + 16 = 0 \\ a n d x^{2} + y^{2} - 14 x - 12 y + 76 = 0 \end{array}$

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New answer posted

2 months ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of a circle of radius 5 which is touching another circle $x^{2} + y^{2} - 2 x - 4 y - 20 = 0$ at (5, 5).

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n c i r c l e i s = (3, - 1) \\ x^{2} + y^{2} - 2 x - 4 y - 20 = 0 \\ 2 g = - 2 \Rightarrow g = - 1 \\ 2 f = - 4 \Rightarrow f = - 2 \\ ∴ C e n t r e C_{1} = (1, 2) \\ a n d r a d i u s r = \sqrt[]{g^{2} + f^{2} - c} \\ = \sqrt[]{1 + 4 + 20} = 5 \\ L e t t h e c e n t r e o f t h e r e q u i r e d c i r c l e b e (h, k) . \\ C l e a r l y, P i s t h e m i d o f C_{1} C_{2} \\ ∴ 5 = \frac{1 + h}{2} \Rightarrow h = 9 \\ a n d 5 = \frac{2 + k}{2} \Rightarrow k = 8 \\ R a d i u s o f t h e r e q u i r e d c i r c l e = 5 \\ ∴ E q u a t i o n o f t h e c i r c l e i s \\ {(x - 9)}^{2} + {(y - 8)}^{2} = {(5)}^{2} \\ \Rightarrow x^{2} + 81 - 18 x + y^{2} + 64 - 16 y = 25 \\ \Rightarrow x^{2} + y^{2} - 18 x - 16 y + 145 - 25 = 0 \\ \Rightarrow x^{2} + y^{2} - 18 x - 16 y + 120 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} + y^{2} - 18 x - 16 y + 120 = 0 . \end{array}$

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