First slide

Hyperbola in conic sections

Question

$If P is a point on the hyperbola x^{2} - y^{2} = a^{2}, C is its centre and S, S^{'} are two foci, then$ $S P \cdot S' P =$

Difficult

A

$a^{2}$
B

${(C P)}^{2}$
C

${(C S)}^{2}$
D

${(S S')}^{2}$

Solution

$x^{2} - y^{2} = a^{2}, C (0, 0)$
$e = \sqrt{2}$
$P (a \sec θ, a \tan θ)$
$s (\sqrt{2} a, 0)$
$s^{1} (- \sqrt{2} a, 0)$
$S P = \sqrt{{(a \sec θ - a \sqrt{2})}^{2} + {(a \tan θ)}^{2}}$
$S P = \sqrt{a^{2} (\sec^{2} θ + 2 - 2 \sqrt{2} \sec θ + \tan^{2} θ)}$
$= a \sqrt{1 + 2 \tan^{2} θ + 2 - 2 \sqrt{2} \sec θ}$
$S P = a \sqrt{(3 + 2 \tan^{2} θ) - 2 \sqrt{2} \sec θ}$
$S^{'} P = \sqrt{(a \sec θ + a \sqrt{2})^{2} + (a \tan θ)^{2}}$
$= \sqrt{\sec^{2} θ + \tan^{2} θ + 2 + 2 \sqrt{2} \sec θ}$
$Step-III: S P S^{1} P = a^{2} \sqrt{9 + 4 \tan^{4} θ + 12 \tan^{2} θ - 8 \sec^{2} θ}$
$= a^{2} \sqrt{4 \tan^{4} θ + 9 + 4 \tan^{2} θ - 8 (\sec^{2} θ - \tan^{2} θ)}$
$= a^{2} \sqrt{{(2 \tan^{2} θ + 1)}^{2}}$
$S P S^{1} P = a^{2} (1 + 2 \tan^{2} θ) \dots \dots (1)$
$Step-IV: C P = \sqrt{(a \sec θ)^{2} + (a \tan θ)^{2}}$
$= a \sqrt{\sec^{2} θ + \tan^{2} θ}$
$= a \sqrt{1 + 2 \tan^{2} θ}$
$C P^{2} = a^{2} (1 + 2 \tan^{2} θ) \dots \dots \dots (2)$
$From (\underline{1) &} (2) S P S^{1} P = (C P)^{2}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App