Question

The line `3x-4y = 12` is a tangent to the ellipse with foci (-2, 3) and (-1, 0). Find the eccentricity of the ellipse.

Answer 1

`f1(-2,3) & f2(-1,0)`
So, `2ae=sqrt((-2+1)^2+ (3-0)^2)=sqrt10` `4a^2e^2=10`
`a^2*(1-b^2/a^2)=5/2`
So, `a^2-b^2=5/2`
Equation of tangent,`y=(3x)/4-3`
&
`y=mx pmsqrt(a^2m^2 +b^2)`
Hence, `m=3/4 & a^2(3/4)^2+ b^2=9` Therefore, `a^2(1+9/16)=9+5/2` `a^2=186/25`
Hence, `4*186/25*e^2=10`
So, `e=sqrt(125/368)=(5sqrt(5))/(4sqrt(23)`