First slide

Hyperbola in conic sections

Question

Statement 1: A normal to the hyperbola with eccentricity 3, meets the transverse axis and conjugate axis at $P$ and $Q$ respectively. The locus of the mid-point of
$P Q$ is a hyperbola with eccentricity $\frac{3}{2 \sqrt{2}}$
Statement 2: Eccentricity of the hyperbol $8 x^{2} - y^{2} =$ $8 a^{2} is \frac{3}{2 \sqrt{2}}$ .

Moderate

A

STATEMENT-1 is True, STATEMENT-2 is True;
STATEMENT-2 is a correct explanation for STATEMENT-1
B

STATEMENT-1 is True, STATEMENT-2 is True;
STATEMENT-2 is NOT a correct explanation for STATEMENT-1
C

STATEMENT-1 is True, STATEMENT-2 is False
D

STATEMENT-1 is False, STATEMENT-2 is True

Solution

Equation of the normal at $(a \sec θ, b \tan θ)$ to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is
$\frac{a x}{\sec θ} + \frac{b y}{\tan θ} = a^{2} + b^{2}$
It meets the axes at $L (\frac{a^{2} + b^{2}}{a \cos θ}, 0), M (0, \frac{a^{2} + b^{2}}{b \cot θ})$
$If (h, k) is the mid-point of L M$
$then h = \frac{a^{2} + b^{2}}{2 a \cos θ}, k = \frac{a^{2} + b^{2}}{2 b \cot θ}$
$\Rightarrow \sec θ = \frac{2 h}{9 a}, \tan θ = \frac{2 b k}{9 a^{2}} [∵ a^{2} + b^{2} = a^{2} e^{2} = 9 a^{2}]$
$Eliminating θ, we get$
$\frac{4 h^{2}}{81 a^{2}} - \frac{4 b^{2} k^{2}}{81 a^{4}} = 1$
Locus of $(h, k) is \frac{x^{2}}{\frac{81 a^{2}}{4}} - \frac{y^{2}}{\frac{81 a^{4}}{4 b^{2}}} = 1$
which is a hyperbola and its eccentricity
$\sqrt{1 + \frac{81 a^{4}}{4 b^{2}} \times \frac{4}{81 a^{2}}} = \sqrt{1 + \frac{a^{2}}{b^{2}}} = \sqrt{1 + \frac{1}{e^{2} - 1}} = \sqrt{1 + \frac{1}{8}}$
$= \frac{3}{2 \sqrt{2}}$
So statement-1 is true. In statement-2, the eccentricity is
$\sqrt{1 + \frac{8 a^{2}}{a^{2}}} = 3$ and the statement-2 is false

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