Conic Sections

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Let $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, a > 0, b > 0,$ be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is $4 (2 \sqrt[]{2} + \sqrt[]{14}) .$ If the eccentricity H is $\frac{\sqrt[]{11}}{2},$ then the value of a² + b² is equal to………….

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Given 2a + 2b = $4 (2 \sqrt[]{2} + \sqrt[]{14})$

$= 4 \sqrt[]{2} (2 + \sqrt[]{7})$

$b^{2} = a^{2} (\frac{11}{4} - 1)$

4b2 = 7a2

b = $\frac{\sqrt[]{7}}{2} a$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Let the tangent to the circle C₁ : x² + y² = 2 at the point M(1, 1) intersect the circle $C_{2} : {(x - 3)}^{2}$ + ${(y - 2)}^{2} = 5,$ at two distinct points A and B. If the tangents to C₂at the points A and B intersect at N, then the area of the triangle ANB is equal to:

21 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Tangent to C₁ at (-1, 1) is T = 0 &nb

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Let PQ be a focal chord of the parabola y2 = 4x such that it subtends an angle of $\frac{π}{2}$ at the point (3, 0). Let the line segment PQ be also a focal chord of the ellipse E: $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1,$ $a^{2} > b^{2} .$ If e is the eccentricity of the ellipse E, then the value of $\frac{1}{e^{2}}$ is equal to:

16 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$A P ⊥ B P$

M₁ M₂ = 1

$\frac{2 t}{t^{2} - 3} \times \frac{2 / 6}{3 - \frac{1}{t^{2}}} = - 1$

t = 1

So, A (1, 2) and B (1, 2) they must be end pts of focal chord.

Length of latus rectum $= \frac{2 b^{2}}{a}$

$4 = \frac{2 b^{2}}{a}$

b2 = 2a and ae = 1

Eccentricity of ellipse (Horizontal)

b²= a² (1 – e²)

2a = a² (1 – e²)

2 = $\frac{1}{e} (1 - e^{2})$

e²+ 2e – 1 = 0

$e = \frac{- 2 \pm \sqrt[]{4 + 4}}{2}$

$e = - 1 + \sqrt[]{2}$

now $\frac{1}{e^{2}}$ $=$ $3$ $+$ $2$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

If the line x – 1 = 0 is a directrix of the hyperbola kx² – y² = 6, then the hyperbola passes through the point

14 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Given hyperbola : $\frac{k x^{2}}{6} - \frac{y^{2}}{6} = 1$ so eccentricity e = $\sqrt[]{1 + k}$ and directrices $x = \pm \frac{a}{e}$

$\Rightarrow x = \pm \frac{\sqrt[]{6}}{\sqrt[]{k} \sqrt[]{k + 1}} \Rightarrow \frac{\sqrt[]{6}}{\sqrt[]{k} \sqrt[]{k + 1}} = 1$

k = 2 therefore equation of hyperbola is $\frac{x^{2}}{3} - \frac{y^{2}}{6} = 1$

hence it passes through the point $(\sqrt[]{5}, - 2)$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Let the abscissae of two points P and Q on a circle be the roots of x² – 4x – 6 = 0 and the ordinates of P and Q be the roots of y² + 2y – 7 = 0. If PQ is a diameter of the circle x² + y² + 2ax + 2by + c = 0, then the value of (a + b +) is……………

8 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Abscissae of PQ are roots of x² – 4x – 6 = 0

Ordinates of PQ are roots of y² + 2y – 7 = 0 and PQ is diameter

$∴$ Equation of circle is $x^{2} + y^{2} - 4 x + 2 y - 13 = 0$ ……………. (i)

But, given $x^{2} + y^{2} + 2 a x + 2 b y + c = 0$ ……………. (ii)

By comparison a = -2, b = 1, c = -13 Þ a + b – c = -2 + 1 + 13 = 12

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

The equation of a common tangent to the parabolas y = x² and y = -(x – 2)²is

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Equation of tangent of slope m to y = x² is y = mx $\frac{1}{4} m^{2}$ - …………. (i)

Equation of tangent of slope m to y = - (x - 2)² is y = m (x – 2) + $\frac{1}{4} m^{2}$ …………… (ii)

If both equation represent the same line therefore on comparing (i) and (ii) we get m = 0, 4

therefore equation of tangent is y = 4x – 4

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

The acute angle between the pair of tangents drawn to the ellipse 2x² + 3y² = 5 from the point (1, 3) is

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$\frac{x^{2}}{{(\sqrt[]{\frac{5}{2}})}^{2}} + \frac{y^{2}}{{(\sqrt[]{\frac{5}{3}})}^{2}} = 1$

Equation of tangent having slope m is

$y = m x \pm \sqrt[]{\frac{5}{3} m^{2} + \frac{5}{3}},$ which passes through (1, 3) and we get m₁ + m₂ = -4 and m₁m₂ = $\frac{44}{9}$

$∴$ Acute angle between the tangents is α = tan^-1 $| \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} | = t a n^{- 1} (\frac{24}{7 \sqrt[]{5}})$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Two tangent lines $l_{1} a n d l_{2}$ are drawn from the point (2, 0) to the parabola 2y² = x. If the lines $l_{1} a n d l_{2}$ are also tangent to the circle (x – 5)² + y²= r, then 17r is equal to……….

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$y^{2} = \frac{- x}{2}$

$y = m x - \frac{1}{8 m}$

This tangent pass through (2, 0)

$m = \pm \frac{1}{4} i . e ., o n e t a n g e n t i s x - 4 y - 2 = 0, 17 r = 9$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Let the tangents at the points P and Q on the ellipse $\frac{x^{2}}{2} + \frac{y^{2}}{4} = 1$ meet at the point $R (\sqrt[]{2}, 2 \sqrt[]{2} - 2) .$ If S is the focus of the ellipse on its negative major axis, then SP² + SQ² is equal to………

15 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Ellipse is $\frac{x^{2}}{4} + \frac{y^{2}}{4} = 1 e = \frac{1}{\sqrt[]{2}} S \equiv (0, - \sqrt[]{2})$

Chord of contact is

$\frac{x}{\sqrt[]{2}} + \frac{(2 \sqrt[]{2} - 2) y}{4} = 1$

$P \equiv (1, \sqrt[]{2}) Q \equiv \sqrt[]{2}, 0$

$∴ {(S P)}^{2} + {(S Q)}^{2} = 13$

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New answer posted•a year ago

Conic Sections Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Let the hyperbola H : $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ pass through the point $(2 \sqrt[]{2}, - 2 \sqrt[]{2}) .$ A parabola is drawn whose focus is same as the focus of H with positive abscissa and the directrix of the parabola passes through the other focus of H. If the length of the latus rectum of the parabola is e times the length of the latus rectum of H, where e is the eccentricity of H, then which of the following points lies on the parabola?

16 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

Foci : S (ae, 0), S’ (ae, 0)

Focus of parabola is (ae, 0)

Now, semi latus rectum of parabola = |SS’| = 2ae

$G i v e n, 4 a e = e (\frac{2 b^{2}}{a})$

B² = 2a²……… (i)

$G i v e n, (2 \sqrt[]{2}, - 2 \sqrt[]{2})$ lies on H

$\Rightarrow \frac{1}{a^{2}} - \frac{1}{b^{2}} = \frac{1}{8} . . . . . . . . (i i)$

From (i) and (ii)

$a^{2} = 4, b^{2} = 8$

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

$∴ e = \sqrt[]{3}$

$\Rightarrow$ equation of parabola is y² $= 8 \sqrt[]{3} x$

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