Let P(3,2) and Q(3,2), where +=2, be two distinct points on the hyperbola x^29-y^24=1.

Question

Let $P(3\sec	heta,2	an	heta)$ and $Q(3\sec\varphi,2	an\varphi)$, where $	heta+\varphi=\dfrac\pi2$, be two distinct points on the hyperbola $\dfrac{x^2}{9}-\dfrac{y^2}{4}=1$. Then the ordinate of the point of intersection of the normals at $P$ and $Q$ is:

Accepted Answer

The normal to $\dfrac{x^2}{9}-\dfrac{y^2}{4}=1$ at $(3\sec\theta,2\tan\theta)$ is
$$3x\cos\theta+2y\cot\theta=13.$$
Similarly at $Q$ (using $\varphi=\tfrac\pi2-\theta$): $3x\sin\theta+2y\tan\theta=13.$

Let $(h,k)$ be the intersection. Subtracting the two normal equations and simplifying (using $\sin\theta-\cos\theta\ne0$) leads to
$$3h=\frac{-2k(\sin\theta+\cos\theta)}{\sin\theta\cos\theta}.$$
Substituting back into one normal equation and simplifying gives $4k\tan\theta-2k=13$ reducing to $-2k=13$, so
$$k=-\frac{13}{2}.$$

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