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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

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  • $\begingroup$ If "$P$ (forming an angle of $\pi/4$ with the major axis)", then the coordinates of $P$ is $(2/\sqrt 5,2/\sqrt 5)$, not $(2\cos\frac{\pi}{4},\sin\frac{\pi}{4})$. If "$P$ whose eccentric angle is $45^\circ$", then the coordinates of $P$ is $(2\cos\frac{\pi}{4},\sin\frac{\pi}{4})$. $\endgroup$ Commented Jun 22, 2021 at 11:48
  • $\begingroup$ I have edited the question. $\endgroup$ Commented Jun 22, 2021 at 12:21
  • $\begingroup$ Then, the diagram does not show what the question describes since the angle $45^\circ$ in the diagram is not the eccentric angle (anomaly) of $P$. $\endgroup$ Commented Jun 22, 2021 at 12:39
  • $\begingroup$ sorry, Just Ignore the diagram $\endgroup$ Commented Jun 22, 2021 at 12:46

2 Answers 2

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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.

enter image description here

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 $m + n = 19$.

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A better approach to this question would be, to take two perpendicular chords from √2 , 1/√2.

diagram(not able to embed it)

At the first thought you would PA and PB as the perpendicular chords and get an equation of line AB: x+2y=0

The other line to take might seem confusing at first but give it a thought,

The tangent at any point is the limiting case of a chord, so if we take point P and a point just next to it, the tangent is the chord. The perpendicular chord would obviously be the normal.

So the two points should be the point just next to it and the other end of its normal chord, the equation of the points joining is again the normal, So technically, we can say point of intersection of normal at that point and the AB is the answer.

Solving should give you coordinates of Q(a,b) and you should get to the final answer of m+n=19

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  • $\begingroup$ bruh why do I get downvotes all the time, this is honestly a good solution instead of adopting a long approach, i cant do the step by step solution, because I don't even know how to type proper math in this, please tell me why you don't like this answer so i can improve. Is it the different approach you don't like because it is not there in internet ? $\endgroup$ Commented May 15, 2023 at 1:19

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