Determine the equations for all lines tangent to the ellipse
$$ f(x,y)=x^2+xy+y^2=1 $$
and passing through the point (0,2)
I'm a bit stuck on this question.
I was thinking if I have an unknown point on the curve called (a.b). Then I can plug it in the curve.
$$f(a.b)=a^2+ab+b^2=1$$
And if I also find the derivative of it, which is:
$$(2a+b,a+2b) $$
But from here I really don't know how to proceed. Maybe if I can somehow set it up as simultaneous equations and find the unknown point, from there I can easily find the tangent line.
Best Answer
Here is one approach:
$x^2+xy+y^2=1$
Taking derivative wrt. $x$, $2x + y + x y' + 2y y' = 0 \implies y' = -\frac{2x+y}{x+2y}$
This is the slope of the tangent line at a point $T (x, y)$ on ellipse. As the tangent line goes through point $P (0, 2)$, we can equate slope of $PT$ to $y'$.
$-\frac{2x+y}{x+2y} = \frac{y-2}{x} \implies x^2+xy+y^2 = x + 2y$.
As point $T (x, y)$ is on ellipse, $x^2 + xy + y^2 = 1$
So, $x = 1 - 2y$ and plugging it into equation of ellipse,
$(1-2y)^2 + y(1-2y) + y^2 = 1 \implies y(y-1) = 0$
That gives us two points on the ellipse $(1, 0)$ and $(-1, 1)$. Tangent lines at these points pass through point $P(0, 2)$.
You can now find the equation of tangent lines.