The following question came from James Stewart's Essential Calculus, problem #42 on page 127. It was asked by one of students, Riley Taylor in my MAT 167 Calculus 1 class. I did not have time to go over details in class, so I am posting a detailed solution here.

*Question:* Find the equations of both the tangent lines to the ellipse $x^2+4y^2=36$ that pass through the point $(12,3)$.

*Solution:* Let $(a,b)$ be a point on the ellipse at which one of the tangent lines through $(12,3)$ meets with the ellipse. To find the slope $\frac{dy}{dx}$ of then tangent line, differentiate $x^2+4y^2=36$ with respect to $x$:

$$2x+8y\frac{dy}{dx}=0$$ or $$\frac{dy}{dx}=-\frac{x}{4y}.$$ Hence, at $(a,b)$ the slope is given by

$$\left[\frac{dy}{dx}\right]_{(a,b)}=-\frac{a}{4b}.$$ The equation of the tangent line is then

$$y-b=-\frac{a}{4b}(x-a).$$ Since the line passes through $(12,3)$,

$$3-b=-\frac{a}{4b}(12-a)$$ or

$$a^2-12a+4b(b-3)=0.$$ Since $(a,b)$ also satisfies the equation of the ellipse $x^2+4y^2=36$, we have

$$a^2+4b^2=36$$ or

$$4b^2=36-a^2.$$ Hence the equation $a^2-12a+4b(b-3)=0$ turns into $a-3+b=0$ or $b=3-a$. Plug this into $a^2+4b^2=36$ for $b$. Then we obtain a quadratic equation:

$$5a^2-24a=0.$$ This equation has two solutions $a=0,\frac{24}{5}$. Therefore we found two points $(a,b)$:

$$(a,b)=(0,3),\ \left(\frac{24}{5},-\frac{9}{5}\right)$$ and the equations of tangent lines

$$y=3,\ y=\frac{2}{3}x-\frac{25}{5}.$$ The following figure shows the ellipse, both the tangent lines and the point $(12,3)$: