SATHEE: Hyperbola 3 Question 1

Hyperbola 3 Question 1

1.

If $x = 9$ is the chord of contact of the hyperbola $x^{2} - y^{2} = 9$ , then the equation of the corresponding pair of tangents is

$(1999, 2 M)$

(a) $9 x^{2} - 8 y^{2} + 18 x - 9 = 0$

(b) $9 x^{2} - 8 y^{2} - 18 x + 9 = 0$

(d) $9 x^{2} - 8 y^{2} + 18 x + 9 = 0$

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Answer:

Correct Answer: 1. (b)

Solution:

Let $(h, k)$ be a point whose chord of contact with respect to hyperbola $x^{2} - y^{2} = 9$ is $x = 9$ .

We know that, chord of contact of $(h, k)$ with respect to hyperbola $x^{2} - y^{2} = 9$ is $T = 0$ .

$\Rightarrow h \cdot x + k (- y) - 9 = 0$

$∴ h x - k y - 9 = 0$

But it is the equation of the line $x = 9$ .

This is possible when $h = 1, k = 0$ (by comparing both equations).

Again equation of pair of tangents is

$\begin{array}{lc} T^{2} = S S_{1} \\ \Rightarrow & (x - 9)^{2} = (x^{2} - y^{2} - 9) (1^{2} - 0^{2} - 9) \\ \Rightarrow & x^{2} - 18 x + 81 = (x^{2} - y^{2} - 9) (- 8) \\ \Rightarrow & x^{2} - 18 x + 81 = - 8 x^{2} + 8 y^{2} + 72 \\ \Rightarrow & 9 x^{2} - 8 y^{2} - 18 x + 9 = 0 \end{array}$

Hyperbola 3 Question 1

1.

Answer:

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Hyperbola 3 Question 1

1.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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