The locus of the point of intersection of the | vedclass

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(B) Let the extremities of the normal chord be $P(a \sec 	heta_1, b 	an 	heta_1)$ and $Q(a \sec 	heta_2, b 	an 	heta_2)$.The equation of the tan

Vedclass · Accepted Answer

$\frac{a^4}{x^2} - \frac{b^4}{y^2} = (a^2 + b^2)^2$

Vedclass · Answer

(B) Let the extremities of the normal chord be $P(a \sec 	heta_1, b 	an 	heta_1)$ and $Q(a \sec 	heta_2, b 	an 	heta_2)$.The equation of the tangent at $P$ is $\frac{x \sec 	heta_1}{a} - \frac{y 	an 	heta_1}{b} = 1$.The intersection point $(h, k)$ of the tangents at $P$ and $Q$ is given by $h = a \frac{\cos(\frac{	heta_1 - 	heta_2}{2})}{\cos(\frac{	heta_1 + 	heta_2}{2})}$ and $k = b \frac{\sin(\frac{	heta_1 - 	heta_2}{2})}{\cos(\frac{	heta_1 + 	heta_2}{2})}$.Since $PQ$ is a normal chord,the condition for the normal at $	heta_1$ to pass through $	heta_2$ is $a \sec 	heta_1 	an 	heta_2 + b 	an 	heta_1 \sec 	heta_2 = 0$ (simplified form).By eliminating the parameters $	heta_1$ and $	heta_2$,the locus of the intersection point $(h, k)$ is found to be $\frac{a^4}{x^2} - \frac{b^4}{y^2} = (a^2 + b^2)^2$.

The locus of the point of intersection of the tangents drawn at the extremities of a normal chord of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is

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