Question

The locus of point of intersection of two tangents of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, meeting at right angles is

• A. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=2$
• B. $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$
• C. $x^2+y^2=a^2+b^2$
• D. $\left(x^2+y^2\right)\left(a^2+b^2\right)=1$.

Answer 1

Answer: (C)

The equation of two perpendicular tangents are
$y=mx+\sqrt{a^2{m}^2{b}^2}...(1)$
and $y=-\frac{1}{m}x+\sqrt{a^2{\left(-\frac{1}{m}\right)}^2+b^2}$
or $my=-x+\sqrt{a^2+b^2{m}^2}$
Since both tangents will pass through $T(h,k)$ so
$k-mh=\sqrt{a^2{m}^2+b^2}$ and $mk+h=\sqrt{a^2+b^2{m}^2}$
Squaring and adding
$(k-mh{)}^2+(mk+h{)}^2=a^2{m}^2+b^2+a^2+b^2{m}^2$
or $h^2\left(m^2+1\right)+k^2\left(1+m^2\right)=a^2\left(1+m^2\right)+b^2\left(1+m^2\right)$
Since $1+m^2\ne 0;h^2+k^2=a^2+b^2$.
Thus, the locus of $T(h,k)$ is $x^2+y^2=a^2+b^2$
This is the equation of a circle. This circle is called the DIRECTOR CIRCLE of the elipse.