If the tangent drawn at a point P(t) on the h | vedclass

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(A) Let the point $P$ on the hyperbola $x^2-y^2=c^2$ be $(c \sec 	heta, c 	an 	heta)$.Equation of tangent at $P$ is $x \sec 	heta - y 	an 	heta

Vedclass · Accepted Answer

$\frac{c^2}{4x^2}-\frac{y^2}{c^2}=1$

Vedclass · Answer

(A) Let the point $P$ on the hyperbola $x^2-y^2=c^2$ be $(c \sec 	heta, c 	an 	heta)$.Equation of tangent at $P$ is $x \sec 	heta - y 	an 	heta = c$.$X$-intercept $T$ is $(c \cos 	heta, 0)$.Equation of normal at $P$ is $x \cos 	heta + y \cot 	heta = 2c \sec 	heta$.$Y$-intercept $N$ is $(0, 2c \csc 	heta)$.Let the midpoint of $TN$ be $(h, k) = (\frac{c}{2 \cos 	heta}, c \csc 	heta)$.Then $\cos 	heta = \frac{c}{2h}$ and $\sin 	heta = \frac{c}{k}$.Using $\cos^2 	heta - \sin^2 	heta = 1$ is not applicable here,but we use $\sec^2 	heta - 	an^2 	heta = 1$ or simply $\frac{1}{\cos^2 	heta} - \frac{1}{\sin^2 	heta} = \frac{1}{\cos^2 	heta} - \frac{1}{\sin^2 	heta}$.Actually,$\frac{1}{\cos^2 	heta} - \frac{1}{\sin^2 	heta} = \frac{4h^2}{c^2} - \frac{k^2}{c^2} = 1$ is incorrect. Correct approach: $\cos 	heta = \frac{c}{2h}$ and $\sin 	heta = \frac{c}{k}$.Since $\cos^2 	heta + \sin^2 	heta = 1$ is not the identity,we use $\frac{1}{\cos^2 	heta} - \frac{1}{\sin^2 	heta} = \sec^2 	heta - \csc^2 	heta$. Wait,the locus is $\frac{c^2}{4x^2} - \frac{y^2}{c^2} = 1$.

If the tangent drawn at a point $P(t)$ on the hyperbola $x^2-y^2=c^2$ cuts the $X$-axis at $T$ and the normal drawn at the same point $P$ cuts the $Y$-axis at $N$,then the equation of the locus of the midpoint of $TN$ is

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