Question

Locus of the point of intersection of perpendicular tangents to the ellipse is

• A. $x^2+y^2=125$
• B. $x^2+y^2=150$
• C. $x^2+y^2=200$
• D. None of these

Answer 1

The correct option is A $x^2+y^2=125$

Locus of point of intersection of perpendicular tangent is the director circle of the ellipse

so,

$x^2+y^2=a^2+100$

$\Rightarrow x^2+y^2=100+a^2$

Now we just have to find a ;

Normal at P;

$axsec\theta -bycosec\theta =a^2{-b}^2$

Slope of the line $CP$ and the normal at $P$ are

$m_1=\frac{a}{b}tan\theta$

$m_2=\frac{b}{a}tan\theta$

$tan\varphi =\frac{m_1{-m}_2}{1+m_1{m}_2}=\frac{tan\theta (\frac{a}{b}-b/a)}{1+ta{n}^2\theta }$

$=\frac{a^2-b^2}{ab}\frac{tan\theta }{se{c}^2\theta }$
$=\frac{a^2-b^2}{ab}\sin \theta \cos \theta =\frac{a^2-b^2}{2ab}\sin 2\theta$
Therefore, the greatest value of $tan\varphi =\frac{a^2-b^2}{2ab}\times 1=\frac{a^2-b^2}{2ab}$
Given that $\frac{a^2-b^2}{2ab}=\frac{3}{2}$. Let $\frac{a}{b}=t$
$\Rightarrow t-\frac{1}{t}=\frac{3}{2}$
$\Rightarrow 2t^2-3t-2=0$
$\Rightarrow 2t^2-4t+t-2=0$
$\Rightarrow (2t+1)(t-2)\Rightarrow \frac{a}{b}=2$
$\Rightarrow e=\frac{\sqrt{3}}{2}$
Rectangle inscribed in the ellipse whose one vertex is $\left(acos\theta \right)\left(2bsin\theta \right)=2absin\left(2\theta \right)$
which has maximum value 2ab. Given that $a=10$, then $b=5\Rightarrow$ maximum area is 100.
Locus of intersection point of perpendicular tangents is $x^2+y^2={10}^2+5^2$ or $x^2+y^2=125$ (director circle).