Position of a Point with Respect to a Parabola - Practice Questions & MCQ

Position of a Point and Focal Chord

Position of a Point with a Parabola

$\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ any point in the plane and point $\mathrm{Q}\left(\mathrm{x}_1, \mathrm{y}_2\right)$ is on the parabola

The point P lies outside, on and inside of the parabola according as

$
\begin{array}{ll} 
& P M>,=\text { or }<Q M \\
\Rightarrow & P M^2>,=\text { or }<Q M^2 \\
\Rightarrow & y_1^2>,=\text { or }<y_2^2 \\
\Rightarrow & y_1^2>,=\text { or }<4 a x_1
\end{array}
$

(a) P lies outside the parabola, $\mathrm{y}_1^2-4 \mathrm{ax}_1>0$
(b) P lies on the parabola, $\mathrm{y}_1^2-4 \mathrm{ax}_1=0$
(c) P lies inside the parabola, $\mathrm{y}_1^2-4 \mathrm{ax}_1<0$

Ends of Focal Chord of a Parabola

\mathrm{PQ} \text { is the focal chord of the parabola } y^2=4 a x . \mathrm{PQ} \text { is passing through the focus } \mathrm{S}(\mathrm{a}, 0) \text {. }

Then point, $\mathrm{P} \equiv\left(\mathrm{at}_1^2, 2 \mathrm{at}_1\right)$, and $\mathrm{Q} \equiv\left(\mathrm{at}_2^2, 2 \mathrm{at}{ }_2\right)$
Point P, S and $Q$ are collinear
So, the Slope of PS = Slope of SQ

$
\begin{array}{lc}
\Rightarrow & \frac{2 a t_1-0}{a t_1^2-a}=\frac{0-2 a t_2}{a-a t_2^2} \\
\Rightarrow & \frac{2 t_1}{t_1^2-1}=\frac{2 t_2}{t_2^2-1} \\
\Rightarrow & t_1\left(t_2^2-1\right)=t_2\left(t_1^2-1\right) \\
\Rightarrow & t_1 t_2\left(t_2-t_1\right)+\left(t_2-t_1\right)=0 \\
\Rightarrow & t_2-t_1 \neq 0 \therefore t_1 t_2+1=0 \\
\therefore & t_1 t_2=-1 \text { or } t_2=-\frac{1}{t_1}
\end{array}
$

Which is the required relation
TIP
If one end point of a focal chord is $\left(a t^2, 2 a t\right)$, then other end point of focal chord becomes $\left(\frac{a}{t^2},-\frac{2 a}{t}\right)$