# Equation of both of the tangent lines to the ellipse $x^2 - 12x + y^2 + 7 = 0$ that pass through the origin.

I need help to solve for the equation of two tangent lines to the ellipse $$x^2 -12x+y^2+7=0$$, which pass through the origin. I've tried a variety of methods from searching similar problems on the internet, however, I can't seem to arrive at an answer.

I know from implicit differentiation that the derivative of $$x^2 -12x+y^2+7=0$$ with respect to $$y$$, is defined as:

$$y' = \frac{(6-x)}{y}$$

And, after that, I'm stumped. How do I go from there? Any help will be much appreciated.

• If the lines pass through the origin, their equation has a very special form: $y=mx$. But you already know the slope of this tangent line at a point $(a,b)$ on the ellipse. Oct 2 '18 at 15:40

Given the ellipse $$E$$ and a line passing by the origin $$L$$

$$E\to b^2(x-x_0)^2+a^2(y-y_0)^2-a^2b^2 = 0\\ L\to p = (x,y) = \lambda \vec v = \lambda(1,m)\\$$

we have that $$E\cap L$$ can be solved as follows

$$b^2(\lambda-x_0)^2+a^2(\lambda m-y_0)^2-a^2b^2 = 0\$$

for

$$\lambda = \frac{a^2 m y_0+b^2 x_0\pm a b \sqrt{a^2 m^2+b^2-(y_0-m x_0)^2}}{a^2 m^2+b^2}$$

but at tangency

$$a^2 m^2+b^2-(y_0-m x_0)^2 = 0$$

and solving for $$m$$ we get

$$m = \frac{\pm\sqrt{a^2 \left(y_0^2-b^2\right)+b^2 x_0^2}-x_0 y_0}{a^2-x_0^2}$$

so the tangency points are at

$$p = \lambda_i\vec v_i\ \ \ i \in \{1,2\}$$

• I noticed that you used the set intersection notation. Is there a sort of application of set theory to analytic geometry and calculus? Oct 2 '18 at 18:06
• @clathratus I used it as a short notation for intersection. Oct 2 '18 at 18:14
• Yeah I got that, I was just wondering. Oct 2 '18 at 18:19

First, since the coefficients of the $$x^2$$ and $$y^2$$ terms agree, if the shape the equation defines is an ellipse, it's actually a circle. In any case, you're already close to an answer.

Hint From your equation for the derivative, the tangent line to the circle at $$(x_0, y_0)$$ is $$y_0 (y - y_0) = (6 - x_0) (x - x_0) .$$ If this line passes through the origin, it is satisfied by $$(x, y) = (0, 0)$$, so $$-y_0^2 = x_0^2 - 6 x_0 .$$

• Thank you for the helpful hint! After getting that equation, I think I have to equate it to the original equation? $x^2 - 12x + y^2 + 7 = 0$? Then solve for the values of x?
– Ryan
Oct 5 '18 at 12:17
• It's even a little easier than that: Since $(x_0, y_0)$ is on the circle, that pair satisfies the circle equation: $x_0^2 - 12 x_0 + y_0^2 + 7 = 0$. By rearranging and combining the equations, we get an affine equation in $x_0$, that is, one without quadratic terms. Oct 5 '18 at 15:19

This ellipse is actually a circle $$(x-6)^2+y^2=29$$ and you can find a tangent without derivative. Just write a equation of circle with diameter $$A(6,0)$$ and $$O(0,0)$$ and calculate where it cuts a given circle. Suppose you get points $$B$$ and $$C$$. Then the lines $$OB$$ and $$OC$$ are the tangnts you seek for.

• I assume you meant center $A(6, 0)$. Oct 3 '18 at 12:23
• No. I meant circle wit diameter $AO$
– Aqua
Oct 3 '18 at 12:36