# Thread: Where does the perpendicular line intersect the ellipse again?

1. ## Where does the perpendicular line intersect the ellipse again?

I have a question which asks me to find the equation of a line perpendicular to the ellipse x^2-xy+y^2=3, at the point (-1, 1). I did this by differentiating implicitly, finding the tangent line then the line perpendicular to that. Pretty simple. The equation of the perpendicular line turned out to be f(x)=-x.

The question now is where does this line intersect the ellipse again? I know the answer is (1, -1) by looking at the graph, but I need to justify this mathematically. Anyone know any algebra gymnastics which will lead me to this answer, based only on f(x)=-x, and the point (-1, 1)?

2. Hello, Ares_D1!

A slight change in notation should clear it up . . .

Find the equation of a line perpendicular to the ellipse $\displaystyle x^2-xy+y^2\:=\:3$, at the point (-1, 1).

The equation of the perpendicular line turned out to be: .$\displaystyle {\color{blue}y \:=\:x}$

The question now is where does this line intersect the ellipse again?

We want the intersection of two curves: the ellipse and the line.

Substitute $\displaystyle y = x$ into: .$\displaystyle x^2-xy+y^2\:=\:3$ . . . and solve.

We'll get two points . . . and we already know one of them . . . Got it?

3. The equation of the perpendicular is y= -x, not y= x.