Question

An ellipse with major axis and minor axis as $8$ and $2$ units respectively, slides along the coordinate axes keeping the contact with axes. Then the locus of its foci is

• A. $x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=16$
• B. $x^2+y^2+\frac{4}{x^2}+\frac{4}{y^2}=16$
• C. $x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=64$
• D. $x^2+y^2+\frac{4}{x^2}+\frac{4}{y^2}=64$

Answer 1

The correct option is C $x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=64$


The product of length of perpendicular from foci to any tangent of the ellipse is $b^2$ (square of semi minor axis).
Given that $2a=8,2b=2\Rightarrow a=4,b=1,e=\frac{\sqrt{15}}{4}$
$X$ and $Y$ axes are the tangents to the ellipse
$x_1{x}_2=b^2=1\&y_1{y}_2=b^2=1$
Let
$(x_1,y_1)=(h,k)\Rightarrow (x_2,y_2)=(\frac{1}{h},\frac{1}{k})S{S}^{\prime }=2ae=2\times 4\times \frac{\sqrt{15}}{4}=2\sqrt{15}\Rightarrow S{S}^{\prime 2}=60\Rightarrow (h-\frac{1}{h}{)}^2+(k-\frac{1}{k}{)}^2=64\Rightarrow h^2+k^2+\frac{1}{h^2}+\frac{1}{k^2}=64\Rightarrow x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=64$