A tangent is drawn to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ to cut the ellipse $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$ at the points $P$ and $Q$. If tangents at $P$ and $Q$ to the ellipse $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$ intersect at right angle, then prove that $\frac{a^2}{c^2}+\frac{b^2}{d^2}=1$.

My attempt at a solution:

Since it is told that the tangents at $P$ and $Q$ intersect at right angle, therefore the point of intersection of the tangents must lie on the director circle of the ellipse $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$ , whose equation is $x^2 + y^2 = c^2 + d^2$.

Let $(a\space \cos\theta \space, \space b \space \sin\theta)$ be a point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at which a tangent is drawn. Therefore equation of the tangent would be $\frac{x\space \cos\theta}{a}+\frac{y\space \sin\theta}{b}=1$. Let this tangent be called $PQ$.

Now, let $(c\space \cos\theta_1 \space, \space d \space \sin\theta_1)$ and $(c\space \cos\theta_2 \space, \space d \space \sin\theta_2)$ be points $P$ and $Q$ on the ellipse $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$ where the tangent intersects it.

Let $R$ be the point of intersection of the tangents drawn at $P$ and $Q$. The coordinates of $R$ is given by $(\frac{a \space \cos(\frac{\theta_1 + \theta_2}{2})}{\cos(\frac{\theta_1 - \theta_2}{2})},\frac{b \space \sin(\frac{\theta_1 + \theta_2}{2})}{\cos(\frac{\theta_1 - \theta_2}{2})})$

Now I get 3 equations, as follows:

  1. The point $R$ satisfies the equation of the director circle.
  2. The point $P$ satisfies the equation of the tangent PQ.
  3. The point $Q$ satisfies the equation of the tangent PQ.

Solving these equations, I get something like: $(c^2-d^2)\sin^4(\theta)+(a^2+b^2+d^2-c^2) \sin^2(\theta)-b^2=0$

Now since it is told in the problem only "A tangent is drawn..." I tried to prove the given equation by making the discriminant of the above equation to 0, but I am unable to get the condition. Is there a simpler approach to this? Thanks for any help.

  • $\begingroup$ I suggest you use $ \sin $ instead if $ sin $ in your post for readability. The results is $\sin(x)$ vs. $sin(x)$, the latter looking like the multiplication of three variables $s\,i\,n$. $\endgroup$ – ja72 Nov 6 '16 at 16:53

1) The first ellipse $E_1$ has equation $$F(x,y)=b^2x^2+a^2y^2-a^2b^2=0$$ and its derivative at the point $(x,y)$ is equal to $$-\dfrac{\dfrac{\partial F}{\partial x}}{\dfrac{\partial F}{\partial y}}=\frac{-b^2x}{a^2y}$$

2) Clearly for an ellipse of equation $\dfrac{X^2}{A^2}+\dfrac{Y^2}{B^2}=1$ the vertical $X=A$ and the horizontal $Y=B$ are tangents to the ellipse which intersects making a right angle at the point $(X,Y)=(A,B)$ (we clearly get a rectangle).

3) Take the tangent line at a point $(x_0,y_0)\in E_1$. You get $$y=-\frac{b^2x_0}{a^2y_0}x+\frac{b^2x_0^2+a^2y_0^2}{a^2y_0}=-\frac{b^2x_0}{a^2y_0}x+\frac{b^2}{y_0}$$

4) The crossing points of this line with the axis are $\left(\dfrac{a^2}{x_0},\space0\right)$ and $\left(0,\space\dfrac{b^2}{y_0}\right)$ and these points are the axes of the given second ellipse $E_2$ of equation (See paragraph 2)) $$\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$$ so this equation becomes $$\frac{x_o^2x^2}{a^4}+\frac{y_0^2y^2}{b^4}=1$$

It is clear now that $(x,y)=(a,b)$ is a point of the ellipse $E_2$ (puting this point in $E_2$ we simply verify that $(a,b)\in E_1$


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