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

Question

Let $P(3,3)$ be a point on the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$. If the normal to it at $P$ intersects the $x$-axis at $(9,0)$ and $e$ is its eccentricity, then the ordered pair $(a^2,e^2)$ is equal to:

Accepted Answer

Normal at $(x_1,y_1)$: $\dfrac{a^2x}{x_1}+\dfrac{b^2y}{y_1}=a^2+b^2$. At $y=0$: $x=\dfrac{x_1(a^2+b^2)}{a^2}=9\Rightarrow\dfrac{3(a^2+b^2)}{a^2}=9\Rightarrow b^2=2a^2$.

Through $P(3,3)$: $\dfrac9{a^2}-\dfrac9{2a^2}=1\Rightarrow\dfrac9{2a^2}=1\Rightarrow a^2=\dfrac92,\ b^2=9$.

$$e^2=1+\frac{b^2}{a^2}=1+2=3.$$

Solutions · 1