Question

If the tangents and normals at the extremities of a focal chord of any standard parabola intersect at $S\equiv (x_1,y_1)$ and $R\equiv (x_2,y_2)$ respectively, then

• A. If the equation of parabola is $y^2=\pm 4ax$, then $y_1=y_2$
• B. If the equation of parabola is $x^2=\pm 4ay$, then $x_1=x_2$
• C. the circle with focal chord as diameter will always pass through the points $R$ and $S$
• D. the circle with $RS$ as diameter will always pass through extemities of focal chord

Answer 1

Answer: (A), (B), (D)

the circle with $RS$ as diameter will always pass through extemities of focal chord

Let $P(a{t}_1^2,2a{t}_1)$ and $Q(a{t}_2^2,2a{t}_2)$ be the extremities of a focal chord of the parabola $y^2=4ax$. The tangents at $P$ and $Q$ intersect at $S\equiv (a{t}_1{t}_2,a(t_1+t_2\left)\right)\Rightarrow S\equiv (-a,a(t_1+t_2\left)\right)$.
$[\because PQ$ is a focal chord$]$
$\therefore x_1=-a$ and $y_1=a(t_1+t_2)$

The normals at $P$ and $Q$ intersect at
$R\equiv (2a+a(t_1^2+t_2^2+t_1{t}_2),-a{t}_1{t}_2(t_1+t_2\left)\right)\Rightarrow R\equiv (a+a(t_1^2+t_2^2),a(t_1+t_2\left)\right)$
$\therefore x_2=a+a(t_1^2+t_2^2)$ and $y_2=a(t_1+t_2)$
Clearly, $y_1=y_2$
Similarly it can be proved for the parabola $y^2=-4ax$
and in same way if we consider the parabola of the form $x^2=\pm 4ay$, then we get $x_1=x_2$

Now from above figure in the quadrilateral $PRQS$
$\angle P+\angle Q=\angle S+\angle R=180$
$\therefore PRQS$ will be cyclic quadrilateral