First slide

Hyperbola in conic sections

Question

If the normal at P on the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ meets the transverse axis at $G, S$ is a foci and
$e$ the eccentricity of the hyperbola then $S G : S P$ is equal to

Moderate

A

$a$
B

$b$
C

$e$
D

$1 / e .$

Solution

Equation of the normal at $P (a \sec θ, b \tan θ)$ is
$\frac{a x}{\sec θ} + \frac{b y}{\tan θ} = a^{2} + b^{2}$ Which meets the transverse axis at
$G (\frac{(a^{2} + b^{2}) \sec θ}{a}, 0)$ $S (a e, 0)$ is
$\begin{array}{l} S G = \frac{(a^{2} + b^{2}) \sec θ}{a} - a e \\ S P = e [a \sec θ - \frac{a}{e}] = a [e \sec θ - 1] \\ \frac{S G}{S P} = \frac{[a^{2} + a^{2} (e^{2} - 1)] \sec θ - a^{2} e}{a \times a [e \sec θ - 1]} \\ = \frac{e^{2} \sec θ - e}{e \sec θ - 1} = e . \end{array}$

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