Question

Variable pairs of chords at right angles and drawn through any point $P$ (with eccentric angle $\frac{\pi }{4}$) on the ellipse $\frac{x^2}{4}+y^2=1$, to meet the ellipse at two points say $A$ and $B$. If the line joining $A$ and $B$ passes through a fixed point $Q(a,b)$ such that $a^2+b^2$ has the value equal to $\frac{m}{n}$, where $m,n$ are relatively prime positive integers, find $(m+n)$.

• A. $5$
• B. $6$
• C. $7$
• D. $8$

Answer 1

The correct option is C $7$

$\frac{x^2}{4}+\frac{y^2}{1}=1$

$a=2,b=1$

$P=(a\cos \pi /4,b\sin \pi /4)$

$P=(\sqrt{2},\frac{1}{\sqrt{2}})$

$MAB=\frac{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}{-\sqrt{2}-\sqrt{2}}=\frac{\sqrt{2}}{2\sqrt{2}}=\frac{1}{2}$

equation of $AB$

$y+\frac{1}{\sqrt{2}}=\frac{1}{2}(x-\sqrt{2})$

$y+\frac{1}{\sqrt{2}}=\frac{-x}{2}+\frac{1}{\sqrt{2}}$

$\therefore x+2y=0\to$ coures from $Q(a,b)$

$a+2b=0$

$b=\frac{-a}{2}$

$a^2+b^2=a^2+\frac{a^2}{4}=\frac{5a^2}{4}$

Let $apt.M(a,\frac{-a}{2})$ i.e. on line $x+2y=0$

$\Rightarrow \frac{a^2}{4}+\frac{a^2/4}{4}=1$

$\Rightarrow \frac{2a^2}{4}=1$

$\Rightarrow a^2=2$

$a^2+b^2=\frac{5a^2}{4}=\frac{5\times 2}{4}=\frac{5}{2}=\frac{m}{n}$

$\therefore m+n=7$