Question

Prove that the normal chord at the point whose ordinate is equal to its abscissa subtends a right angle at the focus.

Answer 1

For $y^2=4ax$ when ordinate is equal to abscissa

$y^2=4ay\Rightarrow y=4a,0\Rightarrow x=4a,0$

So the point be $P$ $(4a,4a)$

Any point on parabola is of the form $(a{t}_1^2,2a{t}_1)$

$\Rightarrow 2a{t}_1=4a{t}_1=2$

If the normal intersect the parabola again then

$t_2=-t_1-\frac{2}{t_1}$

$t_2=-3$

Let it intersect again at $Q(a{t}_2^2,2a{t}_2)$

$\Rightarrow Q(9a,-6a)$

Focus of the parabola is $S(a,0)$

Slope of $PS=m_1=\frac{4a-a}{4a-0}=\frac{3}{4}$

Slope of $QS=m_2=\frac{9a-a}{-6a-0}=-\frac{4}{3}$

$m_1\times m_2=\frac{3}{4}\times -\frac{4}{3}=-1$

Hence proved that lines are perpendicular

So the normal chord subtends right angle at the centre