Let P(4,3) be a point on a hyperbola H:x^2a^2-y^2b^2=1.

Question

Let $P(4,3)$ be a point on a hyperbola $H:\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$. If the sum of the focal distances of $P$ is $\dfrac{8\sqrt5}{3}$, and, for $H$, the length of the latus rectum is $l$ and the product of the focal distances of $P$ is $m$, then $9l^2+6m$ is equal to:

Accepted Answer

The sum condition together with $P$ lying on $H$ gives $a^2=\dfrac52,\ b^2=\dfrac53$ (check: $\dfrac{16}{5/2}-\dfrac{9}{5/3}=\dfrac{32}{5}-\dfrac{27}{5}=1$ ).

$$l=\frac{2b^2}{a}=\frac{10/3}{\sqrt{5/2}}=\frac{2\sqrt{10}}{3},\qquad l^2=\frac{40}{9},\qquad 9l^2=40.$$

$$e^2=1+\frac{b^2}{a^2}=1+\frac23=\frac53,\qquad m=e^2(4)^2-a^2=\frac{80}{3}-\frac52=\frac{145}{6},\qquad 6m=145.$$

$9l^2+6m=40+145=185.$

Solutions · 1