Find $m$ such that the line is normal to the given hyperbola

Question

Find $m$ such that $y=mx+\frac{25}{\sqrt3}$ is normal to $$\frac{x^2} {16}-\frac{y^2}9=1$$

How to go about this question. I don't find any clue.

Answer 1

Say it is normal at point $T(a,b)\in\mathcal{H}$, then from the derivative we get $$ {a\over 8} - {2bk\over 9}=0$$ where $$k = y'(a) = -{1\over m}$$

So we have $$k={9a\over 16b}\implies m = -{16b\over 9a}$$

Since $T$ is on this normal we have $$b = -{16b\over 9a} \cdot a +{25\over \sqrt{3}}\implies b= 3\sqrt{3}$$

Plugin in $${a^2\over 16}-{27\over 9} =1\implies a=\pm 8$$

So $\boxed{m= \pm {2\sqrt{3}\over 3}}$