The foci of a hyperbola are (2,0) and its eccentricity is 32.

Question

The foci of a hyperbola are $(\pm2,0)$ and its eccentricity is $\dfrac32$. A tangent, perpendicular to the line $2x+3y=6$, is drawn at a point in the first quadrant on the hyperbola. If the intercepts made by the tangent on the $x$- and $y$-axes are $a$ and $b$ respectively, then $|6a|+|5b|$ is equal to \_\_\_\_\_\_\_.

Answer: 12

Accepted Answer

$ae=2,\ e=\dfrac32\Rightarrow a=\dfrac43$; $b^2=a^2(e^2-1)=\dfrac{16}9\cdot\dfrac54=\dfrac{20}9$.

Tangent perpendicular to $2x+3y=6$ (slope $-2/3$) has slope $m=3/2$: $y=\dfrac32x\pm\sqrt{4-\dfrac{20}9}=\dfrac32x\pm\dfrac43$.

Taking the branch giving positive intercepts: $x$-intercept $a=\dfrac89$, $y$-intercept $b=\dfrac43$.

$$|6a|+|5b|=6\cdot\frac89+5\cdot\frac43=\frac{16}3+\frac{20}3=12.$$

Solutions · 1