Two tangents to the hyperbola $4x^2-y^2=36$ intersect at the point $(0,4)$. Find the coordinates for the points on the hyperbola for this to occur.

Question

The question is two tangents to the hyperbola $4x^2-y^2=36$ intersect at the point $(0,4)$. Find the coordinates for the points on the hyperbola for this to occur.

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My attempt:

Consider the line $y=mx+c$ as it passes through $(0,4)$ then $c=4$ --> Condition 1

Plug in $y=mx+c$

$$ 4x^2-(mx+c)^2=36$$

$$4x^2-36=m^2x^2+2mcx+c^2$$

$$(m^2-4)x^2+(2mc)x^2+c^2+36=0$$

For tangency $b^2-4ac=0$

Hence

$$ 4m^2c^2-4(c^2+36)(m^2-4)=0$$

$$ 16c^2-144m^2+576=0$$

But $c=4$ due to the line condition hence

$$ m = \pm \frac{2\sqrt{13}}{3}$$

So the lines are

$y= \frac{2\sqrt{13}}{3}x + 4$ and $y= -\frac{2\sqrt{13}}{3}x + 4$

How to find the coordinates for the points on the hyperbola for this to occur tho?

Answer 1

You already have $$(m^2-4)x^2+(2mc)x^2+c^2+36=0$$ Solving this gives $$x=\frac{-mc}{m^2-4}=\frac{-4m}{m^2-4}$$

So, for $m=\pm\frac{2\sqrt{13}}{3}$, the coordinates we want are $$\color{red}{\left(\mp\frac{3}{2}\sqrt{13},-9\right)}$$

Answer 2

The hyperbola can be written as $\frac{x^2}{9} - \frac{y^2}{36} = 1$ and let the tangent at $(3\sec\theta, 6\tan \theta)$ pass through $(0,4)$. The equation of the tangent is $\frac{x}{3}\sec\theta - \frac{y}{6}\tan\theta = 1$ and since $(0,4)$ lies on this, we have $\tan\theta = -\frac{3}{2}$. Thus the points are given by $\left(\frac{3}{2}\sqrt{13}, -9\right), \left(-\frac{3}{2}\sqrt{13}, -9\right)$ according as we take $\cos \theta$ as positive or negative.