MCQ
The normal to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{9}=1$ at the point $(8,3 \sqrt{3})$ on it passes through the point
- A$(15,-2 \sqrt{3})$
- B$(9,2 \sqrt{3})$
- ✓$(-1,9 \sqrt{3})$
- D$(-1,6 \sqrt{3})$
$\frac{64}{a^{2}}-\frac{27}{9}=1 \Rightarrow a^{2}=\frac{64}{4}=16$
equation of normal at $(8,3 \sqrt{3})$ :
$\frac{16 x}{8}+\frac{9 y}{3 \sqrt{3}}=16+9$
$2 x+\sqrt{3} y=25$
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