If P is a point on hyperbola 16{x^2} - 9{y^2} = 144 whose foci are {S₁} and {S₂} , then P{S₁}-

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10-2. Parabola, Ellipse, Hyperbola

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If $P$ is a point on the hyperbola $16{x^2} - 9{y^2} = 144$ whose foci are ${S_1}$ and ${S_2}$, then $P{S_1}- P{S_2} = $

A

$4$

B

$6$

C

$8$

D

$12$

Solution

(b) $\frac{{{x^2}}}{{{3^2}}} – \frac{{{y^2}}}{{{4^2}}} = 1$.

Therefore $P{S_1} ≈ P{S_2} = 2(3) = 6$.

Standard 11

Mathematics

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