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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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Asked by Alok Kumar Singh

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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:

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

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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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Asked by Alok Kumar Singh

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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:

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

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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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Asked by Alok Kumar Singh

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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:

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

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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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Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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 \times 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 \times \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 \times \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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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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Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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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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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Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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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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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Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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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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 $x - 2 y + 3 = 0$ .

3 Views|Posted a year ago

Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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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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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Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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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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).

4 Views|Posted a year ago

Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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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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?

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.]

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

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).

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.

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

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

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 $x - 2 y + 3 = 0$ .

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 $y = x - 1$ .

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).