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What is the locus of the point of intersection of tangents at the extremities of the chords of the ellipse $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\;\;$$ subtending to a right angle at its center?

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user12076043
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  • Please show you have worked on the subject, explain where you are stumped... – Jean Marie Apr 03 '20 at 19:04
  • Do you know for example that the tangent to this ellipse in $(x_0,y_0)$ has equation $xx_0/a^2+yy_0/b^2=1$ ? – Jean Marie Apr 03 '20 at 19:11
  • Yes i know what you are saying. Actually i am not arriving at the answer. I know the answer but don't know the solution. Please help me if you can. – user12076043 Apr 03 '20 at 19:31
  • This question is on it’s way to being closed. If you’d like to prevent that, read How to ask a homework question and edit your question to include that information. – amd Apr 03 '20 at 20:08
  • Pole-polar relationships might provide a way to a solution. “Intersection of tangents” always make me think of looking at that. – amd Apr 03 '20 at 20:10
  • Give names ! Let $A_1(x_1,y_1)$ and $A_2(x_2,y_2)$ be the chord's endpoints, How do you express the orthogonality of $OA_1$ and $OA_2$ ? – Jean Marie Apr 03 '20 at 20:50
  • Hint: using results of my post here and also product of roots:

    $$\frac{y_1 y_2}{x_1 x_2}=m_1 m_2=-1$$

    you can find the equation for the pole $(u,v)$.

     – Ng Chung Tak Apr 03 '20 at 22:26

1 Answers1

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Let the endpoints of the chord $(a\cos\theta_1, b\sin\theta_1)$ and $(a\cos\theta_2, b\sin\theta_2)$. Then, the tangent lines at the endpoints are

$$\frac xa\cos\theta_1 + \frac yb\sin\theta_1 = 1\tag 1$$ $$\frac xa\cos\theta_2 + \frac yb\sin\theta_2 = 1\tag 2$$

Given that they also subtend a right angle at the center, their dot-product is zero,

$$a^2\cos\theta_1\cos\theta_2+b^2\sin\theta_1\sin\theta_2=0\tag 3$$

Take (1) - (2), $$\sin\frac{\theta_2-\theta_1}2 \left( \frac xa \sin\frac{\theta_2+\theta_1}2 - \frac yb \cos\frac{\theta_2+\theta_1}2\right) = 0$$

which leads to

$$\cos\frac{\theta_2+\theta_1}2 = \frac{bx}{\sqrt{b^2x^2+a^2y^2}}, \>\>\>\>\> \sin\frac{\theta_2+\theta_1}2 = \frac{ay}{\sqrt{b^2x^2+a^2y^2}} \tag 4$$ Take (1) + (2),

$$\cos\frac{\theta_2-\theta_1}2 \left( \frac xa \cos\frac{\theta_2+\theta_1}2 + \frac yb \sin\frac{\theta_2+\theta_1}2\right) = 1$$

Use (4) to get,

$$\cos\frac{\theta_2-\theta_1}2 = \frac{ab}{\sqrt{b^2x^2+a^2y^2}}\tag 5$$

Next, rewrite (3) as

$$(a^2+b^2)\cos(\theta_2-\theta_1)+(a^2-b^2)\cos(\theta_2+\theta_1)=0$$

Use the double-angle identity together with the results (4) and (5) to obtain the locus,

$$\frac{x^2}{\frac{a^2}{b^2}(a^2+b^2)} + \frac{y^2}{\frac{b^2}{a^2}(a^2+b^2)} =1$$

which is an ellipse.

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