Consider a hyperbola x^2a^2-y^2b^2=1 having one of its foci at P(-3,0).

Question

Consider a hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ having one of its foci at $P(-3,0)$. If the latus rectum through its other focus subtends a right angle at $P$, and $a^2b^2=\alpha\sqrt2-\beta$, $\alpha,\beta\in\mathbb N$, then $\alpha+\beta$ is \_\_\_\_\_\_\_\_.

Answer: 1944

Accepted Answer

$ae=3$ (since $P=(-ae,0)$). The other focus is at $(ae,0)=(3,0)$, with latus-rectum endpoints $(3,\pm b^2/a)$. Subtending a right angle at $P$ means each half makes $45^\circ$ with the axis:

$$\tan45^\circ=\frac{b^2/a}{2ae}=1 \;\Rightarrow\; \frac{b^2}{a}=2ae=6 \quad(i).$$

Also $a^2e^2=9 \Rightarrow a^2+b^2=9$ (ii). Substituting $b^2=6a$: $a^2+6a-9=0 \Rightarrow a=3(\sqrt2-1)$.

$$a^2=9(3-2\sqrt2),\qquad b^2=18(\sqrt2-1).$$

$$a^2b^2=9(3-2\sqrt2)\cdot6\cdot3(\sqrt2-1)=162(5\sqrt2-7)=810\sqrt2-1134.$$

So $\alpha=810,\ \beta=1134$, and $\alpha+\beta=1944$.

Solutions · 1