A tangent to the hyperbola x^24-y^22=1 meets the x-axis at P and the y-axis at Q.

Question

A tangent to the hyperbola $\dfrac{x^2}{4}-\dfrac{y^2}{2}=1$ meets the $x$-axis at $P$ and the $y$-axis at $Q$. Lines $PR$ and $QR$ are drawn such that $OPRQ$ is a rectangle (where $O$ is the origin). Then $R$ lies on:

Accepted Answer

Tangent at $(a\sec\theta,b\tan\theta)$: $\dfrac{x\sec\theta}{a}-\dfrac{y\tan\theta}{b}=1$, giving $P=(a\cos\theta,0)$, $Q=(0,-b\cot\theta)$.

Since $OPRQ$ is a rectangle, $R=(h,k)=(a\cos\theta,-b\cot\theta)$, so $\cos\theta=\dfrac hb$... more precisely $\cos\theta=\dfrac ha$ and $\sin\theta=-\dfrac{bh}{ak}$ (from $\cot\theta=-k/b$). Squaring and adding using $\sin^2\theta+\cos^2\theta=1$:
$$\frac{a^2}{h^2}-\frac{b^2}{k^2}=1.$$
With $a^2=4,b^2=2$: $\dfrac{4}{x^2}-\dfrac{2}{y^2}=1.$

Solutions · 1