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Prove that the locus of a point from which to the circle $x^2+y^2=a^2$ are inclined at an angle $\alpha$ is $(x^2+y^2-2a^2)^2. \tan^2 (\alpha)=4a^2(x^2 +y^2-a^2)$.

My attempt enter image description here

I could not solve the question. However I have tried to make a figure. Please help me with the further process.

Thanks in Advance.

Note: I have posted this question yesterday. But did not get any answer or hints. So, I am re-posting this again.

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    Do you mean that angle $APB = \alpha$? – 2017-01-04
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    @mary cohen, yea $\angle APB=\alpha$. – 2017-01-04
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    Please don’t repost the same question after barely half a day, especially when you haven’t bothered to correct problems that the original question might have had. – 2017-01-04

1 Answers 1

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Let $P=(h,k)$, $A=(a \cos \theta, a\sin \theta)$ and $B=(a \cos \beta, a\sin \beta)$. Observe that the $\angle BOA=\pi - \alpha=\theta - \beta$. Therefore $$ \cos(\theta -\beta) = -\cos\alpha .............. (1) $$ Consider the right angled triangle $\triangle OBP$, since $OP$ is the bisector of $\angle BOA$, therefore $$\cos^2\left(\frac{\theta -\beta}{2}\right)=\frac{OB^2}{OP^2}=\frac{a^2}{h^2+k^2}.$$ Using this in equation (1), we get \begin{align*} 2\cos^2\left(\frac{\theta -\beta}{2}\right)-1 & = -\cos \alpha\\ \frac{2a^2}{h^2+k^2}-1 & =- \cos \alpha\\ \sec \alpha & = \frac{h^2+k^2}{h^2+k^2-2a^2}\\ \tan^2 \alpha +1 & = \left(\frac{h^2+k^2}{h^2+k^2-2a^2}\right)^2\\ \tan^2 \alpha (h^2+k^2-2a^2)^2 & = (h^2+k^2)^2-(h^2+k^2-2a^2)^2\\ \tan^2 \alpha (h^2+k^2-2a^2)^2 & = (4a^2)(h^2+k^2-a^2). \end{align*} Thus the locus is given by $$\tan^2 \alpha (x^2+y^2-2a^2)^2 = (4a^2)(x^2+y^2-a^2).$$

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    How did you get the co ordinates for A and B? – 2017-01-06
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    @user354073 A and B are points on the circle $x^2+y^2=a^2$, so they can be represented in the form used in my solution. Recall parametric representation. – 2017-01-06