Conic Sections

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of a circle whose center is (3, -1) and which cuts off a chord of length 6 units on the line $2 x - 5 y + 18 = 0$ .

[Hint: To determine the radius of the circle, find the perpendicular distance from the center to the given 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 Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : \\ C e n t r e o f t h e c i r c l e = (3, - 1) \\ L e n g t h o f c h o r d A B = 6 u n i t s \\ C P = | \frac{2 \times 3 - 5 \times 1 + 18}{\sqrt[]{{(2)}^{2} + {(- 5)}^{2}}} | \\ = | \frac{6 + 5 + 18}{\sqrt[]{29}} | = \sqrt[]{29} \\ N o w A B = 6 u n i t s . \\ ∴ A P = \frac{1}{2} A B = \frac{1}{2} \times 6 = 3 u n i t s \\ I n Δ C P A, A C^{2} = C P^{2} + A P^{2} \\ = {(\sqrt[]{29})}^{2} + {(3)}^{2} = 29 + 9 = 38 \\ ∴ A C = \sqrt[]{38} \\ S o, t h e r a d i u s o f t h e c i r c l e, r = \sqrt[]{38} \\ ∴ E q u a t i o n o f t h e c i r c l e i s \\ {(x - 3)}^{2} + {(y + 1)}^{2} = {(\sqrt[]{38})}^{2} \\ \Rightarrow {(x - 3)}^{2} + {(y + 1)}^{2} = 38 \\ \Rightarrow x^{2} + 9 - 6 x + y^{2} + 1 + 2 y = 38 \\ \Rightarrow x^{2} + y^{2} - 6 x + 2 y = 28 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} + y^{2} - 6 x + 2 y = 28 . \end{array}$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of the circle which passes through the points (2, 3) and (4, 5), and its center lies on the straight line $y - 4 x + 3 = 0$ .

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 Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t t h e e q u a t i o n o f t h e c i r c l e i s \\ {(x - h)}^{2} + {(y - k)}^{2} = r^{2} \dots (i) \\ I f t h e c i r c l e passes t h r o u g h (2, 3) a n d (4, 5) t h e n \\ {(2 - h)}^{2} + {(3 - k)}^{2} = r^{2} \dots (i i) \\ a n d {(4 - h)}^{2} + {(5 - k)}^{2} = r^{2} \dots (i i i) \\ S u b t r a c t i n g e q n . (i i i) f r o m e q n . (i i) w e h a v e \\ {(2 - h)}^{2} - {(4 - h)}^{2} + {(3 - k)}^{2} - {(5 - k)}^{2} = 0 \\ \Rightarrow 4 + h^{2} - 4 h - 16 - h^{2} + 8 h + 9 + k^{2} - 6 k - 25 - k^{2} + 10 k = 0 \\ \Rightarrow 4 h + 4 k - 28 = 0 \\ \Rightarrow h + k = 7 \dots (i v) \\ , t h e c e n t r e (h, k) l i e s o n t h e l i n e y - 4 x + 3 = 0 \\ t h e n k - 4 h - 3 = 0 \Rightarrow k = 4 h - 3 \\ P u t t i n g t h e v a l u e o f k i n e q n . (i v) w e g e t \\ h + 4 h - 3 = 7 \Rightarrow 5 h = 10 \Rightarrow h = 2 \\ F r o m (i v) w e g e t k = 5 \\ P u t t i n g t h e v a l u e o f h a n d k i n e q n . (i i) w e g e t \\ {(2 - 2)}^{2} + {(3 - 5)}^{2} = r^{2} \Rightarrow r^{2} = 4 \\ S o, t h e e q u a t i o n o f t h e c i r c l e i s \\ {(x - 2)}^{2} + {(y - 5)}^{2} = 4 \\ \Rightarrow x^{2} + 4 - 4 x + y^{2} + 25 - 10 y = 4 \\ \Rightarrow x^{2} + y^{2} - 4 x - 10 y + 25 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} + y^{2} - 4 x - 10 y + 25 = 0 . \end{array}$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

If the lines $2 x - 3 y = 5$ and $3 x - 4 y = 7$ are the diameters of a circle with an area of 154 square units, then obtain the equation of the circle.

13 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} W e k n o w t h a t t h e intersection point o f t h e d i a m e t e r g i v e s t h e c e n t r e o f t h e c i r c l e . \\ G i v e n e q u a t i o n s o f d i a m e t e r s a r e \\ 2 x - 3 y = 5 \dots (i) \\ 3 x - 4 y = 7 \dots (i i) \\ F r o m e q n . (i) w e h a v e x = \frac{5 + 3 y}{2} \dots (i i i) \\ P u t t i n g t h e v a l u e o f x i n e q n . (i i) w e h a v e \\ 3 (\frac{5 + 3 y}{2}) - 4 y = 7 \\ \Rightarrow 15 + 9 y - 8 y = 14 \Rightarrow y = 14 - 15 \Rightarrow y = - 1 \\ N o w f r o m e q n . (i i i) w e h a v e x = \frac{5 + 3 (- 1)}{2} \Rightarrow x = \frac{5 - 3}{2} \Rightarrow x = 1 \\ S o, t h e c e n t r e o f t h e c i r c l e = (1, - 1) \\ G i v e n t h a t a r e a o f t h e c i r c l e = 154 \\ \Rightarrow π r^{2} = 154 \\ \Rightarrow \frac{22}{7} \times r^{2} = 154 \Rightarrow r^{2} = 154 \times \frac{7}{22} \\ \Rightarrow r^{2} = 7 \times 7 \Rightarrow r = 7 \\ S o, t h e e q u a t i o n o f t h e c i r c l e i s \\ {(x - 1)}^{2} + {(y + 1)}^{2} = {(7)}^{2} \\ \Rightarrow x^{2} + 1 - 2 x + y^{2} + 1 + 2 y = 49 \\ \Rightarrow x^{2} + y^{2} - 2 x + 2 y = 47 \\ H e n c e, t h e r e q u i r e d e q u a t i o n o f t h e c i r c l e i s x^{2} + y^{2} - 2 x + 2 y = 47 . \end{array}$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of the hyperbola with eccentricity $\sqrt[]{2}$ and foci at $(\pm 2, 0)$ .

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 Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t e = \frac{3}{2} a n d f o c i = (\pm 2, 0) \\ W e k n o w t h a t f o c i = (\pm a e, 0) \\ ∴ a e = 2 \Rightarrow a \times \frac{3}{2} = 2 \Rightarrow a = \frac{4}{3} \\ \Rightarrow a^{2} = \frac{16}{9} \\ W e k n o w t h a t b^{2} = a^{2} (e^{2} - 1) \\ \Rightarrow b^{2} = \frac{16}{9} (\frac{9}{4} - 1) = \frac{16}{9} \times \frac{5}{4} = \frac{20}{9} \\ S o, t h e e q u a t i o n o f t h e h y p e r b o l a i s \\ \frac{x^{2}}{16 / 9} - \frac{y^{2}}{20 / 9} = 1 \\ \Rightarrow \frac{9 x^{2}}{16} - \frac{9 y^{2}}{20} = 1 \Rightarrow \frac{x^{2}}{4} - \frac{y^{2}}{5} = \frac{4}{9} \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s \frac{x^{2}}{4} - \frac{y^{2}}{5} = \frac{4}{9} . \end{array}$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the eccentricity of the hyperbola $9 y^{2} - 4 x^{2} = 36$ .

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 Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n i s 9 y^{2} - 4 x^{2} = 36 \\ \Rightarrow \frac{y^{2}}{4} - \frac{x^{2}}{9} = 36 \\ C l e a r l y i t i s v e r t i c a l h y p e r b o l a . \\ W h e r e a = 3 a n d b = 2 \\ N o w, b^{2} = a^{2} (e^{2} - 1) \\ \Rightarrow 4 = 9 (e^{2} - 1) \\ \Rightarrow e^{2} - 1 = \frac{4}{9} \Rightarrow e^{2} = 1 + \frac{4}{9} = \frac{13}{9} \\ ∴ e = \frac{\sqrt[]{13}}{3} \\ H e n c e, t h e r e q u i r e d v a l u e o f e i s \frac{\sqrt[]{13}}{3} . \end{array}$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

If the distance between the foci of a hyperbola is 16 and its eccentricity $\sqrt[]{2}$ is, then obtain the equation of the hyperbola.

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 Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} E q u a t i o n o f h y p e r b o l a i s \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \\ Distance b e t w e e n t h e f o c i = 2 a e \\ 2 a e = 16 \Rightarrow a e = 8 \\ \Rightarrow a \times \sqrt[]{2} = 8 \Rightarrow a = \frac{8}{\sqrt[]{2}} = 4 \sqrt[]{2} [? e = \sqrt[]{2}] \\ N o w, b^{2} = a^{2} (e^{2} - 1) [F o r h y p e r b o l a] \\ \Rightarrow b^{2} = {(4 \sqrt[]{2})}^{2} (2 - 1) \\ \Rightarrow b^{2} = 32 \\ a = 4 \sqrt[]{2} \Rightarrow a^{2} = 32 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s \frac{x^{2}}{32} - \frac{y^{2}}{32} = 1 \Rightarrow x^{2} - y^{2} = 32 . \end{array}$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

If the line $y = m x + 1$ is tangent to the parabola $y^{2} = 4 x$ , then find the value of $m$ .
[Hint: Solving the equation of the line and parabola, obtain a quadratic equation and apply the tangency condition to find $m$ ].

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 Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y^{2} = 4 x \dots (i) \\ a n d y = m x + 1 \dots (i i) \\ F r o m e q n . (i) a n d (i i) w e g e t \\ {(m x + 1)}^{2} = 4 x \\ \Rightarrow m^{2} x^{2} + 1 + 2 m x - 4 x = 0 \\ \Rightarrow m^{2} x^{2} + (2 m - 4) x + 1 = 0 \\ A p p l y i n g c o n d i t i o n o f tangency w e h a v e \\ {(2 m - 4)}^{2} - 4 m^{2} \times 1 = 0 \\ \Rightarrow 4 m^{2} + 16 - 16 m - 4 m^{2} = 0 \\ \Rightarrow - 16 m = - 16 \Rightarrow m = 1 \\ H e n c e, t h e r e q u i r e d v a l u e o f m i s 1 . \end{array}$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

If the points (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then find the equation of the parabola.

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n t h a t : \\ V e r t e x = (0, 4) a n d F o c u s = (0, 2) \\ L e t P (x, y) b e a n y point o n t h e p a r a b o l a . P B i s p e r p e n d i c u l a r t o t h e d i r e c t r i x . \\ A c c o r d i n g t o t h e d e f i n i t i o n o f p a r a b o l a, w e h a v e \\ P F = P B \\ \Rightarrow \sqrt[]{{(x - 0)}^{2} + {(y - 2)}^{2}} = | \frac{0 + y - 6}{\sqrt[]{0 + 1}} | [? E q u a t i o n o f d i r e c t r i x i s y = 6] \\ \Rightarrow \sqrt[]{x^{2} + {(y - 2)}^{2}} = (y - 6) \\ S q u a r i n g b o t h s i d e s, w e h a v e \\ x^{2} + y^{2} + 4 - 4 y = y^{2} + 36 - 12 y \\ \Rightarrow x^{2} - 4 y + 12 y - 32 = 0 \\ \Rightarrow x^{2} + 8 y - 32 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} + 8 y = 32 . \end{array}$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the length of the line segment joining the vertex of the parabola $y^{2} = 4 a x$ and a point on the parabola where the line segment makes an angle $θ$ to the x-axis.

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 Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n p a r a b o l a i s y^{2} = 4 a x \\ L e t P (a t^{2}, 2 a t) b e a n y o n t h e p a r a b o l a . \\ I n Δ P O A, \\ t a n θ = \frac{2 a t}{a t^{2}} = \frac{2}{t} \\ \Rightarrow t = \frac{2}{t a n θ} \\ \Rightarrow t = 2 c o t θ \dots (i) \\ O P = \sqrt[]{{(a t^{2} - 0)}^{2} + {(2 a t - 0)}^{2}} \\ = \sqrt[]{a^{2} t^{4} + 4 a^{2} t^{2}} \\ = a t \sqrt[]{t^{2} + 4} \\ = a \times 2 c o t θ \sqrt[]{4^{} + 4} [? t = 2 c o t θ] \\ = 2 a c o t θ . 2 \sqrt[]{c o t^{2} θ + 1} = 4 a c o t θ c o s e c θ \\ = 4 a . \frac{c o s θ}{s i n θ} . \frac{1}{s i n θ} = \frac{4 a c o s θ}{s i n^{2} θ} \\ H e n c e, t h e r e q u i r e d l e n g t h = \frac{4 a c o s θ}{s i n^{2} θ} . \end{array}$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the coordinates of a point on the parabola $y^{2} = 8 x$ whose focal distance is 4.

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 Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n p a r a b o l a i s y^{2} = 8 x \dots (i) \\ C o m p a r i n g w i t h t h e e q u a t i o n o f p a r a b o l a y^{2} = 4 a x \\ 4 a = 8 \Rightarrow a = 2 \\ N o w f o c a lDistance = | x + a | \\ \Rightarrow | x + a | = 4 \\ \Rightarrow (x + a) = \pm 4 \Rightarrow x + 2 = \pm 4 \\ \Rightarrow x = 4 - 2 = 2 a n d x = - 6 \\ B u t x \neq - 6 ∴ x = 2 \\ P u t x = 2 i n e q u a t i o n (i) w e g e t \\ y^{2} = 8 \times 2 = 16 \\ ∴ y = \pm 4 \\ S o, t h e c o o r d i n a t e s o f t h e points a r e (2, 4), (2, - 4) . \\ H e n c e, t h e r e q u i r e d c o o r d i n a t e s a r e (2, 4), (2, - 4) . \end{array}$

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