Find equation of two tangent lines to ellipse $x^2+4y^2=36$ drawn from $(12,3)$

Question

Find the equation of the two tangent lines to the ellipse $x^2+4y^2=36$ that pass through the point $(12,3)$.

I tried using implicit differentiation, and then I didn't know where to go from there.

Answer 1

Well one of the tangents is obvious:

Answer 2

Use the general equation below for the pair of tangents drawn from the point $(x_1,y_1)$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$,

$$(\frac{x^2}{a^2} +\frac{y^2}{b^2}-1)(\frac{x_1^2}{a^2} +\frac{y_1^2}{b^2}-1) =(\frac{x_1x}{a^2} +\frac{y_1y}{b^2}-1)^2$$

Thus, with $(x_1,y_1)=(12,3)$, $a=6$ and $b=3$, the tangent-line equation is $$x^2+4y^2-36=(x+y-3)^2$$

or, in its factorized form to show the two tangent lines explicitly,

$$(y-3)(2x-3y-15)=0$$

Answer 3

Sorry for the hand written solutions, so we find $m$ and we put the value of $m$ in the tangent equation to get the line