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

Question

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

Finding Slope

The derivative of $x^2 + 4y^2 = 36$ is $y'= -\dfrac{x}{4y}$.

Finding arbitrary point where tangent line is at

If I arrange the equation $x^2 + 4y^2 = 36$, $$y= \pm\frac{\sqrt{x^2+36}}{2}$$

Let $(k, \pm\frac{\sqrt{k^2+36}}{2})$ be the point where tangent line is at. Thus the tangent line slope at that point is $$-\frac{k}{4\left(\pm\frac{\sqrt{k^2+36}}{2}\right)}$$

Equation of tangent line is

$y-y_0 = m (x-x_0)$

Let $(k, \pm\frac{\sqrt{k^2+36}}{2})$ be $(x_0, y_0)$. Let $(12,3)$ be $(x,y)$

$$12-k = -\frac{k}{4\left(\pm\frac{\sqrt{k^2+36}}{2}\right)}\left(3- \pm \frac{\sqrt{k^2+36}}{2}\right)$$

And I'm stuck here. It gets really confusing?

Answer 1

The equation of a line that pass through the point $(12,3)$ is $(y-3)=m(x-12)$. So the system: $$ \begin{cases} (y-3)=m(x-12)\\ x^2+4y^2=36 \end{cases} $$

represents the common points of the line and the ellipse and the line is a tangent iff this system has only one solution .

Solving the first equation for $y$ and substituting in the second equation we find a second degree equation in $x$ that contains the parameter $m$
$$ P)^{(2)}(x,m)=0 $$ take the discriminat of this equation that is a second degree polynomial in $m$ : $\Delta(m)$.

Solving the equation $\Delta(m)=0$ you find the values of $m$ such that the system has only one solution , i.e. the slopes of the two ( if the point is external to the ellipse) tangent lines .

Note that this method does not use the derivative (but the fact that the equation of a conic section is second degree) and is, in general, simpler.

Answer 2

$y=mx+n$ is an equation of the tangent to ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ iff $n^2=a^2m^2+b^2$.

Since $$\frac{x^2}{36}+\frac{y^2}{9}=1$$ is an equation of our ellipse, we get $a^2=36$, $b^2=9$ and it's enough to solve the following system: $$n^2=36m^2+9$$ $$12m+n=3,$$ which gives $m=0$, $n=3$ and $y=3$ or

$m=\frac{2}{3}$, $n=-5$ and $y=\frac{2}{3}x-5.$

Done!

Answer 3

To continue the way you started, I suggest backing up a step. You’ve determined that the slope of the tangent at the point $(x_0,y_0)$ on the ellipse is $-\frac{x_0}{4y_0}$, so the equation of the tangent line at that point is $(y-y_0)=-\frac{x_0}{4y_0}(x-x_0)$, or $x_0x+4y_0y=36$. We want this line to pass through $(12,3)$, so we must have $12x_0+12y_0=36$, i.e., the points at which the tangent lines intersect the circle lie on the line $x+y=3$. This line is known as the polar line of the point $(12,3)$ with respect to this ellipse. The problem then becomes that of finding the intersection of the polar line with the ellipse, i.e., of solving the system of equations $$x^2+4y^2=36 \\ x+y=3.$$ I trust that you’re able to do that and then construct the corresponding equations of the tangent lines.