Question

The locus of the image of the focus of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ with respect to any of the tangents to the ellipse is

• A. $(x+4{)}^2+y^2=100$
• B. $(x+2{)}^2+y^2=50$
• C. $(x-4{)}^2+y^2=100$
• D. $(x-2{)}^2+y^2=50$

Answer 1

Answer: (A), (C)

$(x-4{)}^2+y^2=100$


Equation of the ellipse is
$\frac{x^2}{25}+\frac{y^2}{9}=1$
Foci $(\pm 4,0)$
Let $S^{\prime }(h,k)$ be the image.
Mid point of $S{S}^{\prime }$ is $M\equiv (\frac{h\pm 4}{2},\frac{k}{2})$
$S{S}^{\prime }$ cuts tangent line at point $M$ which lies on the auxiliary circle of the ellipse , since foot of perpendicular from foci upon any tangent lies on auxilary circle.
Auxiliary circle of the ellipse is $x^2+y^2=25$
$\therefore M$ lies on auxiliary circle.
So,
$\Rightarrow {\left(\frac{h\pm 4}{2}\right)}^2+\frac{k^2}{4}=25$
$\therefore$ Required locus equation is $(x\pm 4{)}^2+y^2=100$