Question

An ellipse with major axis $4$ and minor axis $2$ touches both the coordinate axes, then locus of its focus is

• A. $(x^2-y^2)(1+x^2{y}^2)=-16x^2{y}^2$
• B. $(x^2-y^2)(1+x^2{y}^2)=16x^2{y}^2$
• C. $(x^2+y^2)(1+x^2{y}^2)=16x^2{y}^2$
• D. $(x^2-y^2)(1-x^2{y}^2)=16x^2{y}^2$

Answer 1

Answer: (C)

$(x^2+y^2)(1+x^2{y}^2)=16x^2{y}^2$

Fix the ellipse as $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and let p and q be the signed perpendicular distances from $(ae,0)$ to a pair of mutually perpendicular tangents. Then the focus is $(p,q)$ referred to the coordinate system where the tangents are the axes.

Given tangent,

$y=m_{1}x\pm \sqrt{b^2+a^2{m}_1^2}$

then ,

$p=\pm \frac{-m_{1}ae\pm \sqrt{b^2+a^2{m}_1^2}}{\sqrt{1+m_1^2}}$

So,

${p\left(\sqrt{1+m_1^2}\right)\pm m_{1}ae}^2=b^2+a^2{m}_1^2$

$\Rightarrow p^2\pm \frac{2m_{1}aep}{\sqrt{1+m_1^2}}-b^2=0$

Rearranging this gives,

$\frac{1}{m_1^2}=\frac{4a^2{e}^2{p}^2}{p^2-b^2}-1$

If $m_2$ is gradient of the other tangent then similarly ...

$\frac{1}{m_2^2}=\frac{4a^2{e}^2{p}^2}{q^2-b^2}-1$

Applying condition $m_1{m}_2=-1$ :

$[\frac{4a^2{e}^2{p}^2}{p^2-b^2}-1]\times [\frac{4a^2{e}^2{p}^2}{q^2-b^2}-1]=1$

$\Rightarrow 16a^4{e}^4{p}^2{q}^2-4a^2{e}^2{p}^2(q^2-b^2{)}^2-4a^2{e}^2{q}^2(p^2-b^2{)}^2=0$

$a=2,b=1\Rightarrow 12p^2{q}^2-p^2(q^2-1{)}^2-q^2(p^2-1{)}^2=0\Rightarrow 16p^2{q}^2-p^2(q^4+1)-q^2(p^4+1)=0$

$\Rightarrow 16p^2{q}^2=(p^2+q^2)(1+p^2{q}^2)$

this is locus relative to the perpendicular tangents.

Replacing $(ae,0)$ by $(-ae,0)$ leads to the same result.

Replacing $(p,q)$ by $(x,y)$ we get,

$\Rightarrow (x^2+y^2)(1+x^2{y}^2)=16x^2{y}^2$