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  • A circle meets the x-axis at A(-a, 0) and B(a, 0). <img alt

A circle \mathrm{x^2+y^2=a^2} meets the x-axis at A(-a, 0) and B(a, 0). \mathrm{P(\alpha)} and \mathrm{Q(\beta)} are two points on the circle so that \mathrm{\alpha-\beta=2 \gamma}, where \mathrm{\gamma} is a constant. Then locus of the point of intersection of AP and BQ is.

Option: 1

\mathrm{x^2+y^2-2 a y ~\tan ~\gamma=a^2}


Option: 2

\mathrm{x^2+y^2-2 a y ~\cot ~\gamma=a^2}


Option: 3

\mathrm{x^2+y^2+2 a y ~\tan ~\gamma=a^2}


Option: 4

\mathrm{x^2+y^2-a y ~\tan ~\gamma=a^2}


Answers (1)

best_answer

Let,

The coordinates of A are (-a, 0) and that of P are \mathrm{(a \cos \alpha, a \sin \alpha)}.
Therefore, the equation of AP is
           \mathrm{ y =\frac{a \sin \alpha}{a(\cos \alpha+1)}(x+a) }
\mathrm{ \Rightarrow \quad y =\tan \frac{\alpha}{2}(x+a) }         .....(1)

Similarly, the equation of BQ is
      \mathrm{ y =\frac{a \sin \beta}{a(\cos \beta-1)}(x-a) \\ }
\mathrm{ \Rightarrow y =-\cot \frac{\beta}{2}(x-a) }         .....(2)

We now eliminate \mathrm{ \alpha } and \mathrm{ \beta } from (1) and (2). Therefore,
\mathrm{ \tan \frac{\alpha}{2}=\frac{y}{a+x}, \tan \frac{\beta}{2}=\frac{a-x}{y} }
Now, \mathrm{ \alpha-\beta=2 \gamma \Rightarrow \gamma=\frac{\alpha}{2}-\frac{\beta}{2} }

\mathrm{ \therefore \tan \gamma=\frac{\tan \frac{\alpha}{2}-\tan \frac{\beta}{2}}{1+\tan \frac{\alpha}{2} \tan \frac{\beta}{2}}=\frac{\frac{y}{a+x}-\frac{a-x}{y}}{1+\frac{y}{a+x} \cdot \frac{a-x}{y}} \\ }
\mathrm{ \Rightarrow \tan \gamma=\frac{y^2-\left(a^2-x^2\right)}{(a+x) y+(a-x) y}=\frac{x^2+y^2-a^2}{2 a y} \\ }
\mathrm{ \Rightarrow x^2+y^2-2 a y \tan \gamma=a^2 }
 

Posted by

Divya Prakash Singh

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