# Finding a tangent to an ellipse parallel to a given line

Problem: Find the lines that are tangent to the ellipse $x^2 + 4y^2 = 8$ and parallel to $x +2y = 6$.

I tried to find the derivative of $x^2 + 4y^2 = 8$ and I got: $$\frac{dx}{dy} = -\frac{x}{2y}.$$ Not quite sure if it's right, but I tried to equate it with the slope of $x +2y = 6$, which is $-\frac{1}{2}$. So I get, $$-\frac{1}{2} = -\frac{x}{2y}$$ Now, I'm stuck and I don't know what to do. How do I get the tangent lines? I have no knowledge of any point of that tangent line, only the slope. I appreciate any help. Thanks.

• It should be $x=2y$. – Tunk-Fey Apr 18 '14 at 12:52
• How did you compute the derivative? It looks quite wrong to me. I get $2xdx + 8ydy = 0$, so $\frac{dx}{dy} = -\frac{4y}{x}$. Perhaps you meant $\frac{dy}{dx}$, which would be $-\frac{x}{4y}$. – TonyK Apr 18 '14 at 18:51

To find the points of intersection, note that $$-\frac 12 = -\frac{x}{2y} \iff x = y$$

This means that the two lines parallel to the given line are tangent to the eclipse when $x = y$. We can find the two points of tangency to the ellipse, $\;(x_1, y_1), (x_2, y_2)\;$ where $\;x_1 = y_1,\;$ $x_2 = y_2\;$ by using the fact that $x = y$, together with the equation of the ellipse:

$$x=y,\quad x^2 + 4y^2 = 8$$ $$\implies x^2 + 4x^2 = 8 \iff 5x^2 = 8 \iff x_{1, 2} = \pm \sqrt{\frac{8}{5}} \iff y_{1, 2} = \pm \sqrt{\frac 85}$$

Now you have the desired slope of both lines, $m = -1/2$, and a point on each each of your desired lines, $(x_1, y_1), (x_2, y_2)$, so you can use the point-slope form of a line to construct: $$l_1:\quad y - y_i = m(x - x_i)$$ for each $i = 1, i = 2$

• Thank you so much. Your answer helped me a lot. – rainbowbutterunikitteh Apr 18 '14 at 12:47
• You're very welcome, Dio Paulo! – Namaste Apr 18 '14 at 12:49

Rewrite the equation of the ellipse $x^2+4y^2=8$ as \begin{align} x^2+4y^2&=8\\ 4y^2&=8-x^2\\ y^2&=\frac{8-x^2}{4}\\ y&=\left(\frac{8-x^2}{4}\right)^{\frac12}.\tag1 \end{align}

The slope of the lines that are tangent to the ellipse is \begin{align} \frac{dy}{dx}&=\frac{d}{dx}\left(\frac{8-x^2}{4}\right)^{\frac12}\\ &=\frac12\left(\frac{8-x^2}{4}\right)^{-\frac12}\left(\frac{-2x}{4}\right)\\ &=\frac12\left(\frac{4}{8-x^2}\right)^{\frac12}\left(-\frac{1}{2}x\right)\\ &=-\frac{1}{2}x\left(\frac{1}{8-x^2}\right)^{\frac12} \end{align} or the quickest way $$\frac d{dx}(x^2+4y^2=8)\quad\Rightarrow\quad 2x+8y \frac {dy}{dx}=0\quad\Rightarrow\quad \frac {dy}{dx}=-\frac {x}{4y}.$$ Since it is parallel with the line $x +2y = 6$, then \begin{align} -\frac{1}{2}x\left(\frac{1}{8-x^2}\right)^{\frac12}&=-\frac{1}{2}\\ x\left(\frac{1}{8-x^2}\right)^{\frac12}&=1.\tag2 \end{align} Square both sides of $(2)$, yield \begin{align} \frac{x^2}{8-x^2}&=1\\ x^2&=8-x^2\\ 2x^2&=8\\ x_{1,2}&=\pm2\tag3 \end{align} $y$ can be found by plugging in $(3)$ to $(1)$, yield \begin{align} y_{1,2}&=\left(\frac{8-2^2}{4}\right)^{\frac12}\\ &=\pm1, \end{align} where $(x_{1,2},y_{1,2})$ are the points of contact. Now, the equation of tangent line can be found by using $$\frac{y-y_{1,2}}{x-x_{1,2}}=-\frac12.$$

So if $(a,b)$ be the point of contact , you have $$-\frac12=-\frac a{2b}\iff a=b\text{ and }a^2+4b^2=8$$

Solve for $a,b$

So, we have $$\frac{y-b}{x-a}=-\frac12$$