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