Locus of the poles of focal chord of a parabola is
- the axis
- a focal chord
- the directrix
- the tangent at the vertex
Solution
$ \begin{aligned} y-2 a t_1 & =\frac{2 a t_2-2 a t_1}{a t_2^2-a t_1^2}\left(x-a t_1^2\right) \\ \Rightarrow \quad\left(t_2+t_1\right)\left(y-2 a t_1\right) & =2\left(x-a t_1^2\right) \\ \Rightarrow \quad\left(t_1+t_2\right) y-2 x & =2 a t_1 t_2 \end{aligned} $ This line is passing through $(a, 0)$ $ \begin{aligned} \Rightarrow & & \left(t_1+t_2\right)\left(-2 a t_1\right)=\frac{2}{t_1+t_2}\left(a-a t_1^2\right) \\ \Rightarrow & & t_1 t_2=-1 \end{aligned} $ Let $P\left(x_1, y_1\right)$ be the pole of (i) w.r.t. $y^2=4 a x$ Its polar is $y y_1=2 a\left(x+x_1\right)$ From equaiton (i) and (iii), we get $ \frac{t_1+t_2}{y_1}=\frac{1}{a}=\frac{2 a t_1 t_2}{2 a x_1} $ From last two relations, we get $ \begin{array}{ll} & x_1=a t_1 t_2 \\ \Rightarrow & x_1=-a \\ \therefore \text { locus is } x=-a & \end{array} $
Asked in: AP EAMCET 2002