Question

Match the column
The parabola $y^2=4ax$ has a chord $AB$ joining points $A(a{t}_1^2,2a{t}_1)$ and $B(a{t}_2^2,2a{t}_2)$.

Column - I	Column - II
(A) $AB$ is a normal chord if	(P) $t_2=-t_1-\frac{1}{2}$
(B) $AB$ is a focal chord	(Q) $t_2=\frac{-4}{t_2}$
(C) $AB$ subtends ${90}^{\circ }$ at point $(0,0)$ if	(R) $t_2=\frac{-1}{t_1}$ (S) $t_2=-t_1-\frac{2}{t_1}$

Answer 1

Given

$y^2=4ax$

$A=(a{t}_1^2,2a{t}_1)$

$B=(a{t}_2^2,2a{t}_2)$

By the property If normal at point $P(a{t}_1^2,2a{t}_1)$ meets the parabola at $Q(a{t}_2^2,2a{t}_2)$

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

So, $AB$ is a normal chord if

$\left(S\right)-t_2=-t_1-\frac{-2}{t_1}$

By the property - For a chord joining $P$ & $Q$ and passing through focus then $t_1{t}_2=-1$

So, $AB$ is a focal chord is

$\left(R\right)-t_2=\frac{-1}{t_1}$

By the property - If ends of a chord subtend a right angle at the vertex of the parabola then $m_1{m}_2=-1$, So, $t_2=\frac{-4}{t_1}$

So, $AB$ subtends ${90}^o$ at point $(0,0)$ if $\left(Q\right)-t_2=\frac{-4}{t_1}$