First slide

Hyperbola in conic sections

Question

The locus of the middle points of the normal chords of the rectangular hyperbola $x^{2} - y^{2} = a^{2}$ is

Moderate

A

${(y^{2} - x^{2})}^{3} = 4 a^{2} x^{2} y^{2}$
B

${(y^{2} - x^{2})}^{2} = 4 a^{2} x^{2} y^{2}$
C

${(y^{2} + x^{2})}^{3} = 4 a^{2} x^{2} y^{2}$
D

${(y^{2} + x^{2})}^{2} = 4 a^{2} x^{2} y^{2}$

Solution

Equation of the normal at $(a \sec θ, a \tan θ)$ to the hyperbola $x^{2} - y^{2} = a^{2}$ is
$\frac{a x}{\sec θ} + \frac{a y}{\tan θ} = a^{2} + a^{2} = 2 a^{2} (1)$
Let $(h, k)$ be the middle point of this normal then its equation is
$h x - k y - a^{2} = h^{2} - k^{2} - a^{2} (T = S_{1})$
$\Rightarrow h x - k y = h^{2} - k (2)$
Comparing (1) and (2) we get
$\begin{array}{l} h \sec θ = - k \tan θ = \frac{h^{2} - k^{2}}{2 a} \\ \Rightarrow \sec θ = \frac{h^{2} - k^{2}}{2 a h}, \tan θ = \frac{h^{2} - k^{2}}{- 2 a k} \\ So 1 = \sec^{2} θ - \tan^{2} θ = \frac{{(h^{2} - k^{2})}^{2}}{4 a^{2}} (\frac{1}{h^{2}} - \frac{1}{k^{2}}) \\ \Rightarrow {(h^{2} - k^{2})}^{3} + 4 a^{2} h^{2} k^{2} = 0 \\ Locus of (h, k) is {(y^{2} - x^{2})}^{3} = 4 a^{2} x^{2} y^{2} \end{array}$

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