SATHEE: Hyperbola 2 Question 9

Hyperbola 2 Question 9

9.

Let $P (a \sec θ, b \tan θ)$ and $Q (a \sec φ, b \tan φ)$ , where $θ + φ = \frac{π}{2}$ , be two points on the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ .

If $(h, k)$ is the point of the intersection of the normals at $P$ and $Q$ , then $k$ is equal to

$(1999, 2 M)$

(a) $\frac{a^{2} + b^{2}}{a}$

(b) $- \frac{a^{2} + b^{2}}{a}$

(d) $- \frac{a^{2} + b^{2}}{b}$

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Answer:

Correct Answer: 9. (d)

Solution:

Firstly, we obtain the slope of normal to $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ at $(a \sec θ, b \tan θ)$ . On differentiating w.r.t. $x$ , we get

$\frac{2 x}{a^{2}} - \frac{2 y}{b^{2}} \times \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{b^{2}}{a^{2}} \frac{x}{y}$

Slope for normal at the point ( $a \sec θ, b \tan θ$ ) will be

$- \frac{a^{2} b \tan θ}{b^{2} a \sec θ} = - \frac{a}{b} \sin θ$

$∴$ Equation of normal at $(a \sec θ, b \tan θ)$ is

$\begin{aligned} y - b \tan θ = - \frac{a}{b} \sin θ (x - a \sec θ) \\ \Rightarrow (a \sin θ) x + b y = (a^{2} + b^{2}) \tan θ \\ \Rightarrow a x + b cosec θ = (a^{2} + b^{2}) \sec θ \end{aligned}$

Similarly, equation of normal to $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ at

$(a \sec φ, b \tan φ)$ is $a x + b y cosec φ = (a^{2} + b^{2}) \sec φ$

On subtracting Eq. (ii) from Eq. (i), we get

$b (cosec θ - cosec φ) y = (a^{2} + b^{2}) (\sec θ - \sec φ)$

$\Rightarrow y = \frac{a^{2} + b^{2}}{b} \cdot \frac{\sec θ - \sec φ}{cosec θ - cosec φ}$

But $\frac{\sec θ - \sec φ}{cosec θ - cosec φ} = \frac{\sec θ - \sec (π / 2 - θ)}{cosec θ - cosec (π / 2 - θ)}$

$= \frac{\sec θ - cosec θ}{\sec θ - \sec θ} = - 1$

$y = - \frac{a^{2} + b^{2}}{b}, i.e. k = - \frac{a^{2} + b^{2}}{b}$

Hyperbola 2 Question 9

9.

Answer:

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Hyperbola 2 Question 9

9.

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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