The locus of the point of intersection of tan | vedclass

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(C) Let $P(h, k)$ be the point of intersection of tangents at the ends of a normal chord of the hyperbola $x^2 - y^2 = a^2$. The equation of the chord

Vedclass · Accepted Answer

$a^2(y^2 - x^2) = 4 x^2 y^2$

Vedclass · Answer

(C) Let $P(h, k)$ be the point of intersection of tangents at the ends of a normal chord of the hyperbola $x^2 - y^2 = a^2$. The equation of the chord of contact for point $P(h, k)$ is $hx - ky = a^2$ ... $(i)$. Since this is a normal chord of the hyperbola $x^2 - y^2 = a^2$,its equation must be of the form $ax \sec 	heta + ay 	an 	heta = a^2 + b^2$. For $x^2 - y^2 = a^2$,$a=b$,so the normal is $x \sec 	heta + y 	an 	heta = 2a$ ... (ii). Comparing equations $(i)$ and (ii),we have $\frac{h}{\sec 	heta} = \frac{-k}{	an 	heta} = \frac{a^2}{2a} = \frac{a}{2}$. Thus,$\sec 	heta = \frac{2h}{a}$ and $	an 	heta = \frac{-2k}{a}$. Using the identity $\sec^2 	heta - 	an^2 	heta = 1$,we get $(\frac{2h}{a})^2 - (\frac{-2k}{a})^2 = 1$. $\frac{4h^2}{a^2} - \frac{4k^2}{a^2} = 1 \Rightarrow 4(h^2 - k^2) = a^2$. Wait,re-evaluating the normal chord equation for $x^2 - y^2 = a^2$: The normal at point $(a \sec \phi, a 	an \phi)$ is $x \cos \phi + y \cot \phi = 2a$. The chord of contact is $hx - ky = a^2$. Comparing $\frac{h}{\cos \phi} = \frac{-k}{\cot \phi} = \frac{a^2}{2a} = \frac{a}{2}$. $\cos \phi = \frac{2h}{a}$ and $\cot \phi = \frac{-2k}{a}$. Since $\csc^2 \phi - \cot^2 \phi = 1$,we have $(\frac{a}{2h})^2 - (\frac{-2k}{a})^2 = 1$. $\frac{a^2}{4h^2} - \frac{4k^2}{a^2} = 1 \Rightarrow a^4 - 16h^2k^2 = 4a^2h^2$. Actually,the standard result for the locus of intersection of tangents at the ends of a normal chord of $x^2 - y^2 = a^2$ is $a^2(y^2 - x^2) = 4x^2y^2$.

The locus of the point of intersection of tangents at the ends of a normal chord of the hyperbola $x^2 - y^2 = a^2$ is

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