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

Question

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.

Thank you in advance!

Answer 1

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

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

Answer 2

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 $$-y_0^2 = x_0^2 - 6 x_0 .$$

Answer 3

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.