The tangent at an extremity (in the first quadrant) of the latus rectum of the hyperbola x…

Question

The tangent at an extremity (in the first quadrant) of the latus rectum of the hyperbola $\dfrac{x^2}{4}-\dfrac{y^2}{5}=1$ meets the $x$-axis and $y$-axis at $A$ and $B$ respectively. Then $(OA)^2-(OB)^2$, where $O$ is the origin, equals:

Accepted Answer

Here $a^2=4,\ b^2=5\Rightarrow e^2=1+	frac54=	frac94$. The latus-rectum end point is $\left(ae,\dfrac{b^2}{a}ight)=\left(3,\dfrac52ight)$.

Tangent at $(x_1,y_1)$: $\dfrac{xx_1}{4}-\dfrac{yy_1}{5}=1\Rightarrow\dfrac{3x}{4}-\dfrac{y}{2}=1\Rightarrow \dfrac{x}{4/3}+\dfrac{y}{-2}=1.$

So $OA=	frac43,\ OB=-2$. Then $OA^2-OB^2=\dfrac{16}{9}-4=-\dfrac{20}{9}.$

Solutions · 1