>Find the Equation of ellipse that is tangent to the lines $x-4y=10$ and $x+y=5$

Question

Find the Equation of Ellipse that is tangent to the lines $x-4y=10 $ and $x+y=5$

I don't know what is the idea of this problem , how can i begin

Thank you for your help

Answer 1

There are infinitly many of ellipses. One of them: $ (x+y)^2+(x-4y-5)^2=25 $.

Solution Method: Let's apply affine transformation $u=x+y$, $v=x-4y$. Therefore lines $x+y=5,x-4y=10$ maps to $u=5,v=10$. In the plane $uv$, we need a circle that tangent to lines $u=5,v=10$. For example $u^2+(v-5)^2=25$. Now, in the $xy$ plane, inverse of the circle is an ellipse and

$$ (x+y)^2+(x-4y-5)^2=25 $$

that tangent to $x+y=5,x-4y=10$.

NOTE: We can check here

Answer 2

The simplest ellipses are circles. You know that the points on the bisector of an angle are equally far away from the sides of the angle. So in step 1, find a bisecting line. Note that since the original equations describe non-parallel lines, you have 4 angles. Once you found one bisector, take any arbitrary point $(x_0,y_0)$ on it, except the intersection. Find the distance $d$ from that point to one of the lines. That is the radius of the circle. The equation of the ellipse that satisfies your conditions is then $$\frac{(x-x_0)^2}{d^2}+\frac{(y-y_0)^2}{d^2}=1$$