Conic Sections

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Let the tangents at two points A and B on the circle x² + y² – 4x + 3 = 0 meet at origin O(0, 0). Then the area of the triangle OAB is:

21 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

C : (x – 2)² + y² = 1

Equation of chord AB : 2x = 3

$O A = O B = \sqrt[]{3}$

$A M = \frac{\sqrt[]{3}}{2}$

$A r e a o f Δ O A B = \frac{1}{2} (2 A M) (O M)$

$= \frac{3 \sqrt[]{3}}{4} s q . u n i t$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Let $l$  be a line which is normal to the curve y = 2x² + x + 2 at a point P on the curve. If the point Q(6, 4) lies on the line $l$  and O is origin, then the area of the triangle OPQ is equal to

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$y = 2 x^{2} + x + 2 . . . . (i)$

$\frac{d y}{d x} = 4 x + 1$

Slope of normal

$- \frac{d x}{d y} = - \frac{1}{4 x + 1}$

Equation of PQ y - β = $- \frac{1}{4 α + 1} (x - α)$

It passes (6, 4)

$\Rightarrow (4 - β) (4 α + 1) = - (6 - α)$

$4 α^{3} + 3 α^{2} - 3 α - 3 = 0$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

If the tangents drawn at the points O(0, 0) and $P (1 + \sqrt[]{5}, 2)$  on the circle x² + y² – 2x – 4y = 0 intersect at the point Q, then the area of the triangle OPQ is equal to

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

x^{2} + y^{2} - 2 x - 4 y = 0

Centre (1, 2) r = $\sqrt[]{5}$

Equation of OQ is x . 0 + y . 0 – (x + 0) – 2 (y + 0) = 0

⇒ x + 2y = 0 ……… (i)

Equation of PQ is $x (1 + \sqrt[]{5}) + 2 y - (| x + 1 + \sqrt[]{5}) - 2 (y + 2) = 0$

Solving (i) & (ii), $Q (\sqrt[]{5} + 1, - \frac{\sqrt[]{5} + 1}{2})$

$= \frac{1}{2} | \frac{6 + 2 \sqrt[]{5} + 4 \sqrt[]{5} = 4}{2} | = \frac{6 \sqrt[]{5} + 10}{4} = \frac{3 \sqrt[]{5} + 5}{2}$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Let the eccentricity of the hyperbola $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 b e \sqrt[]{\frac{5}{2}}$ and length of its latus rectum be $6 \sqrt[]{2}, i f y = 2 x + c$ is a tangent to the hyperbola H, then the value of c²is equal to

7 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$a^{2} (e^{2} - 1) = b^{2}$

$e = \sqrt[]{\frac{5}{2}} \Rightarrow b^{2} = \frac{3 a^{2}}{2}$

Length of latus rectum $\frac{2 b^{2}}{a} = 6 \sqrt[]{2}$

$3 a = 6 \sqrt[]{2} \Rightarrow a = 2 \sqrt[]{2}$

$b = 2 \sqrt[]{3}$

y = 2x + c is tangent to hyperbola

$∴ c^{2} = a^{2} m^{2} - b^{2} = 20$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

If the equation of the parabola, whose vertex is at (5, 4) and the directrix is 3x + y – 29 = 0, is $x^{2} + a y^{2} + b x y + c x + d y + k = 0,$  then a + b + c + d + k is equal to

51 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

Given vertex is (5, 4) and directrix 3x + y – 29 = 0

Let foot of perpendicular of (5, 4) on directrix be (x₁, y₁)

$\frac{x_{1} - 5}{3} = \frac{y_{1} - 4}{1} = \frac{- (- 10)}{10}$

$∴ (x_{1}, y_{1}) = (8, 5)$

So, focus of parabola will be S = (2, 3)

Let P(x, y) be any point

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

The number of points of intersection of $| z - (4 + 3 i) | = 2 a n d | z | + | z - 4 | = 6, z \in C,$  is

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

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

(a) $\frac{x^{2}}{4} - \frac{y^{2}}{4} = 1$
(b) $\frac{x^{2}}{4} - \frac{y^{2}}{5} = 1$
(c) $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$
(d) None of these

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The distance between the foci of a hyperbola is 16, and its eccentricity is 2. Its equation is

(a) $x^{2} - y^{2} = 32$
(b) $\frac{x^{2}}{4} - \frac{y^{2}}{9} = 1$
(c) $2 x - 3 y^{2} = 7$
(d) None of these

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is half the distance between the foci is

(a) $\frac{3}{2}$
(b) $2$
(c) $\frac{3}{2}$
(d) None of these

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e n g t h o f t h e l a t u s r e c t u m o f t h e h y p e r b o l a \\ = \frac{2 b^{2}}{a} = 8 \Rightarrow b^{2} = 4 a \dots (i) \\ Distance b e t w e e n t h e f o c i = 2 a e \\ T r a n s v e r s e a x i s = 2 a \\ a n d C o n j u g a t e a x i s = 2 b \\ ∴ \frac{1}{2} (2 a e) = 2 b \Rightarrow a e = 2 b \Rightarrow b = \frac{a e}{2} \dots (i i) \\ \Rightarrow b^{2} = \frac{a^{2} e^{2}}{4} \Rightarrow 4 a = \frac{a^{2} e^{2}}{4} [F r o m e q n . (i)] \\ \Rightarrow 16 = a e^{2} ∴ a = \frac{16}{e^{2}} \\ N o w b^{2} = a^{2} (e^{2} - 1) \\ \Rightarrow 4 a = a^{2} (e^{2} - 1) \\ \Rightarrow \frac{4}{a} = e^{2} - 1 \Rightarrow \frac{4}{16 / e^{2}} = e^{2} - 1 \\ \Rightarrow \frac{e^{2}}{4} = e^{2} - 1 \Rightarrow e^{2} - \frac{e^{2}}{4} = 1 \Rightarrow \frac{3 e^{2}}{4} = 1 \\ \Rightarrow e^{2} = \frac{4}{3} ∴ e = \frac{2}{\sqrt[]{3}} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

If $e$ is the eccentricity of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ (where $a < b$ ), then

(a) $b^{2} = a^{2} (1 - e^{2})$
(b) $a^{2} = b^{2} (1 - e^{2})$
(c) $b^{2} = a^{2} (e^{2} - 1)$
(d) $a^{2} = b^{2} (e^{2} - 1)$

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n i s \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a < b) \\ ∴ E c c e n t r i c i t y e = \sqrt[]{1 - \frac{a^{2}}{b^{2}}} \Rightarrow e^{2} = 1 - \frac{a^{2}}{b^{2}} \\ \Rightarrow \frac{a^{2}}{b^{2}} = (1 - e^{2}) \Rightarrow a^{2} = b^{2} (1 - e^{2}) \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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