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Find the locus of the foot of perpendicular from the centre of the ellipse $${x^2\over a^2} +{y^2\over b^2} = 1$$ on the chord joining the points whose eccentric angles differ by $π/2.$

My approach is:
Consider two points $P$ and $Q$ such that $P(a\cos\theta , b\sin\theta),\; Q(a\sin\theta, -b\cos\theta).\;$
Using these two points I write the equation of the chord and then I use the formula of the foot of a perpendicular from $\;(X_1,Y_1)\;$ to a given line, to find the locus.
But I can't find the locus because this approach is very difficult and takes a lot of time. So please solve this in a simpler way.

user376343
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Tips
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  • What have you tried? Select two points $P_1$ and $P_2$ on the ellipse, what's the relationship between the coordinates if the eccentric angles difference is $\pi/2$? – Andrei Jan 07 '19 at 16:41
  • My approach is :- Taking two point P and Q, P(acos● , bsin●) and Q (asin●, -bcos●). Using these two point , I write the equation of chord and then I use the Formula of Foot of a perpendicular from (X1,Y1) to a given line, to find the locus. But I can't find the locus because this approach is very difficult and take a lot of time. So plz solve this que in a simple way. – Tips Jan 07 '19 at 17:20
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    OP is correct and you are confusing him. Eccentric angle is exactly the argument in parametric equation. – Vasily Mitch Jan 07 '19 at 18:25
  • @saketkumar: Please edit the body of your Question to include the approach given in your Comment. This will greatly improve the Question and put your observations in a clearer context. – hardmath Jan 08 '19 at 21:21

2 Answers2

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Let the centre of the ellipse be O and the foot be R. Also let point P is $(acos\theta, bsin\theta)$ then Q is $(acos(\theta+\frac{\pi}{2}), bsin(\theta+\frac{\pi}{2})$ or $(-asin\theta, bcos\theta)$ then PQ is

$$\frac{y - bcos\theta}{x + asin\theta} = \frac{b(sin\theta - cos\theta)}{a(sin\theta + cos\theta)} or \frac{\frac{y}{b} - cos\theta}{\frac{x}{a} + sin\theta} = \frac{sin\theta - cos\theta}{sin\theta + cos\theta}$$

Also OR is

$$\frac{y}{x} = - \frac{a(sin\theta + cos\theta)}{b(sin\theta - cos\theta)} or \frac{by}{ax} = -\frac{sin\theta + cos\theta}{sin\theta - cos\theta}$$

From these two equations we have

$$x^2 + y^2 = - axsin\theta + bycos\theta$$

Divide both sides by $\sqrt{a^2x^2 + b^2y^2}$ and let $tan\phi = \frac{by}{ax}$ then we have

$sin(\theta - \phi) = \frac{x^2 + y^2}{\sqrt{a^2x^2 + b^2y^2}}$ or $\theta = sin^{-1}(\frac{x^2 + y^2}{\sqrt{a^2x^2 + b^2y^2}}) + \phi$

Also from the second equation

$$\frac{y}{x} = - \frac{a}{b}\frac{tan\theta + 1}{-1 + tan\theta}$$

Substitute $\theta$ in the last equation using tangent compound angles and simplify, we have the following equation of the locus provided x and y not equal to zero.

$$a^2x^2 + b^2y^2 = 2(x^2 + y^2)^2$$

KY Tang
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  • Why you Divide both sides by √a^2x^2+b^2y^2 and also why you let tan●= by/ax – Tips Jan 08 '19 at 03:44
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    In order to express $\theta$ in terms of x and y I have to define $\phi$ so that sin the sum of angles can be used. The same for tan. = by/ax. This is the only way to solve x and y from the 2 equations. – KY Tang Jan 08 '19 at 23:07
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    The expression is obtained by the sqrt of the sum of squares of the coefficient of sin theta and cos theta in order to make the whole expression equal to sin theta + phi. – KY Tang Jan 08 '19 at 23:49
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The envelop of all your chords is the ellipse itself scaled down by $\sqrt2$. It's easy to see by affine transforming ellipse into circle, drawing all the chords, and transforming it back.

All the chords are now tangents to smaller ellipse, so the locus of perpendicular foot has a special name pedal curve. Ellipse pedal curves have well-known formulae.

If you want to get to the formula yourself, you should really follow your approach. It's not that hard, just a little bit of algebra involved.

Vasily Mitch
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  • Can you solve this question according to my approach because I can't find my mistake. So plz solve for me, I know it is not a correct way to handle the mathematics question but I am very frustrated after solving same question 5 times and don't getting the correct answer. – Tips Jan 07 '19 at 18:48