Find the locus of the mid-points of the chord of the ellipse $b^2 x^2 + a^2 y^2 = a^2 b^2$ which subtends a right angle at the origin

Question

In a book, I found the solution is $(a^2+b^2)(b^2x^2+a^2y^2)^2=a^2b^2(b^4x^2+a^4y^2)$. Here is my solution:

Let $P$ and $Q$ the ends of a ellipse chord $b^2 x^2 + a^2 y^2 = a^2 b^2$ which subtends a right angle at the origin. Let $M(x,y)$ the mid-point of chord $\overline{PQ}$, then $\vec{MP} = \vec{QM} =(u,v)$. Thus:

$\begin{array}{ll} \vec{OP} &= \vec{OM} + \vec{MP} = (x,y)+(u,v)=(x+u,y+v) \implies P(x+u,y+v) \\ \vec{OQ} &= \vec{OM} - \vec{QM} = (x,y)-(u,v)=(x-u,y-v) \implies Q(x-u,y-v) \\ \end{array}$

As $P(x+u,y+v)$ lies on ellipse, then:

$b^2 (x+u)^2 + a^2 (y+v)^2 = a^2 b^2$...(1)

As $Q(x-u,y-v)$ lies on ellipse, then:

$b^2 (x-u)^2 + a^2 (y-v)^2 = a^2 b^2$...(2)

As $\overline{OP} \perp \overline{OQ}$:

$\begin{array}{ll} & \overline{OP} \perp \overline{OQ} \iff (x+u)(x-u)+(y+v)(y-v)=0 \iff \\ & x^2-u^2+y^2-v^2=0 \iff x^2+y^2=u^2+v^2 \end{array}$...(3)

(1)+(2):

$\begin{array}{ll} & \displaystyle b^2 \left( (x+u)^2 + (x-u)^2 \right) + a^2 \left( (y+v)^2 + (y-v)^2 \right) = 2a^2 b^2 \\ & \displaystyle 2b^2 \left( u^2+x^2 \right) + 2a^2 \left( v^2+y^2 \right) = 2a^2b^2 \iff \\ & \displaystyle b^2 \left( u^2+x^2 \right) + a^2 \left( v^2+y^2 \right) = a^2b^2 \end{array}$...(4)

(2)-(1):

$\begin{array}{ll} & \displaystyle b^2 \left( (x+u)^2 - (x-u)^2 \right) + a^2 \left( (y+v)^2 - (y-v)^2 \right) = 0 \\ & \displaystyle b^2(2x)(2u) + a^2(2y)(2v) = 0 \iff b^2xu+a^2yv=0 \\ \end{array}$...(5)

Solving the system with (3) and (5):

$\begin{array}{ll} & \displaystyle \begin{cases} & u^2+v^2 = x^2+y^2 \\ & b^2xu+a^2yv =0 \end{cases} \implies \begin{cases} & u^2+v^2 = x^2+y^2 \\ & u = \displaystyle -\frac{a^2yv}{b^2x} \\ \end{cases} \implies \\ & \displaystyle \frac{a^4y^2v^2}{b^4x^2}+ v^2 = x^2+y^2 \implies a^4y^2v^2+b^4x^2v^2= b^4x^2 \left( x^2+y^2 \right) \implies \\ & \left(a^4y^2+b^4x^2 \right) v^2 =b^4x^2\left( x^2+y^2 \right) \implies v^2 = \displaystyle \frac{b^4x^2\left( x^2+y^2 \right)}{a^4y^2+b^4x^2} \implies \\ & \displaystyle u^2 = x^2+y^2 - \frac{b^4x^2\left( x^2+y^2 \right)}{a^4y^2+b^4x^2} = \frac{ \left( a^4y^2+b^4x^2- b^4x^2 \right) \left( x^2+y^2 \right)}{a^4y^2+b^4x^2} = \frac{ a^4y^2 \left( x^2+y^2 \right)}{a^4y^2+b^4x^2} \end{array}$...(6)

Substituting $u^2$ y $v^2$ from (6) in (4):

$\begin{array}{ll} & \displaystyle b^2 \left( \left( \frac{ a^4y^2 \left( x^2+y^2 \right)}{a^4y^2+b^4x^2} \right)+x^2 \right) + a^2 \left( \left( \frac{b^4x^2\left( x^2+y^2 \right)}{a^4y^2+b^4x^2} \right)+y^2 \right) = a^2b^2 \implies \\ & \displaystyle \left( a^2+b^2 \right) \left( b^2x^2+a^2y^2 \right)^2 = a^2b^2 \left( b^4x^2+a^4y^2 \right) \end{array}$

Answer 1

$$E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.....(1)$$ The equation of chord whose mid point is $(h,k)$ is given by $$C:\frac{hx}{a^2}+\frac{ky}{b^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}=F \text{(let)}.....(2)$$ We homogenize $E$ and with $C$ in $x,y$ as $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=\left(\frac{hx}{a^2F}+\frac{ky}{b^2F}\right)^2.....(3)$$ This is combined equation of lines OP and OQ where O is origin (center of ellipse) and P, Q are points of intersection of the chord with ellipse. OP and OQ will be perpendicular then sum of coefficients of $x^2$ and that of $y^2$ sums to zero: $A+B=0$. From (3) we get $$\frac{h^2}{a^4F^2}+\frac{k^2}{b^4F^2}-\frac{1}{a^2}-\frac{1}{b^2}=0$$ $$\implies \frac{h^2}{a^4}+\frac{k^2}{b^4}=\left(\frac{h^2}{a^2}+\frac{k^2}{b^2}\right)^2\left(\frac{a^2+b^2}{a^2b^2}\right)$$ By changing $x,y \to x,y$, we get the required locus.

Answer 2

The essence of your mistake is that you are confusing parametric coordinates with polar coordinates.

What you have done in your calculations is parametrized the ellipse's points, i.e. $$P(\theta) = (a \cos \theta, b \sin \theta).$$ So for a given angle $\theta$, you get a point $P$ on the ellipse. But this angle is not a polar angle unless $a = b$ and the ellipse is a circle.

To understand why, look at the inverse tangent of the slope, which will give you the actual angle made by the ray $OP$: it is $$\tan^{-1} \frac{y_p}{x_p} = \tan^{-1} \frac{b \sin \theta}{a \cos \theta} = \tan^{-1} \left(\frac{b}{a} \tan \theta\right).$$ And here it is obvious that this will not be $\theta$ unless $a = b$.

Therefore, when you compute $Q$ by simply adding $\pi/2$ to $\theta$, you are not actually getting a perpendicular vector. The actual perpendicular to $P(x_p,y_p)$ is a point $Q(x_q, y_q)$ that satisfies

$$\frac{x_q^2}{a^2} + \frac{y_q^2}{b^2} = 1, \quad x_p x_q + y_p y_q = 0. \tag{1}$$

The first condition restricts $Q$ to lie on the ellipse. The second condition is the perpendicularity criterion, which comes from the dot product: $$\vec{OP} \cdot \vec{OQ} = 0 \iff \vec{OP}\perp \vec{OQ}.$$