Variable pairs of chords at right angles are drawn through a point $P$ (with eccentric angle $\pi/4$) on the ellipse. Variable pairs of chords at right angles are drawn through a point $P$ (with eccentric angle $\pi/4$) 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 fix point $Q (a,b)$ such that $a^2+b^2$ has value equal to $\frac{m}{n}$, where $m,n$ are relatively prime positive integers. Find $(m+n)$
My Approach:
Note:-$m_{AB}$ denotes Slope of $AB$.
Equation of $AB$ is
$\frac{y}{1}\cdot sin \frac {(\alpha + \beta)}{2}+ \frac{x}{2}\cdot cos\frac {(\alpha+ \beta)}{2}= cos\frac {(\alpha -\beta)}{2}$
Let $A=(2cos\alpha,sin\alpha)$ and $B=(2cos\beta,sin\beta)$
$P=(2cos\frac{\pi}{4},sin\frac{\pi}{4})$
$m_{AP}=\frac{sin\alpha - \frac{1}{\sqrt2}}{{2cos\alpha}-\frac{2}{\sqrt2}}$
$m_{BP}=\frac{sin\alpha - \frac{1}{\sqrt2}}{{2cos\alpha}-\frac{2}{\sqrt2}}$
Because $AP$ and $BP$ are perpendicular so $m_{AP}\cdot m_{BP}=-1$
After solving I reach to $\frac{5}{2}cos(\alpha-\beta)+\frac{3}{2}cos(\alpha+\beta)$= $\frac{2cos \frac{\alpha- \beta}{2}}{\sqrt2} \biggl ( sin \frac {(\alpha - \beta)}{2}+cos\frac {(\alpha - \beta)}{2}\biggl)+\frac{5}{2}$
How to get end Result?
Can i get end result using my method or similar to my method?
This Question is same as below but he used some direct result.
Ellipse in which two chords are perpendicular to each other
Prove that the chord of the ellipse passes through a fixed point

My idea would be to simplify the working. The question does not ask us to prove that the chords pass through the same point $Q$. It states that if they do, what is the coordinates of $Q$. So if they do pass through a common point $Q$, its coordinates can be found using any such two chords.

Given eccentric angle of $\frac{\pi}{4}$, coordinates of $P$: $\left(\sqrt2, \dfrac{1}{\sqrt2}\right)$
So if we take the first pair as a horizontal and a vertical line,
Coordinates of $A$: $\left(\sqrt2, -\dfrac{1}{\sqrt2}\right)$
Coordinates of $B$: $\left(-\sqrt2, \dfrac{1}{\sqrt2}\right)$
Equation of line $AB$ turns out to be $x+2y = 0 \ \ $ ...$(i)$
Now you have two approaches you can follow,
you can show using the answer to one of the questions you linked (link) that the normal line at $P$ will pass through point $Q$. That makes it easier to find the coordinates of $Q$.
Or just take point $B'$ as $\left(-\sqrt2, - \dfrac{1}{\sqrt2}\right)$ so slope of line $PB'$ is $ \dfrac{1}{2}$.
Hence equation of line $PA'$ will be,
$\left(y - \dfrac{1}{\sqrt2}\right) = -2 (x - \sqrt2) \implies 2x + y = \dfrac{5}{\sqrt2}$
Plugging it into equation of ellipse, you get the coordinates of $A'$ as $\left(\dfrac{23\sqrt2}{17}, -\dfrac{7}{17\sqrt2} \right)$.
That leads to equation of line $A'B'$ as,
$ y + \dfrac{1}{\sqrt2} = \dfrac{1}{8} (x + \sqrt2) \ \ $ ...$(ii)$
Solving $(i)$ and $(ii)$ should give you coordinates of $Q (a, b)$ and you should get to the final answer of $m + n = 19$.
