Question

Point $P$ lies on $y^2=4ax$ & $N$ is foot of perpendicular from $P$ on its axis. A straight line is drawn parallel to the axis to bisect $NP$ and meets the curve in $Q$. $NQ$ meets the tangent at the vertex in a point $T$ such that $AT=kNP$, then the value of of $k$ is : (where $A$ is the vertex.)

• A. $\frac{3}{2}$
• B. $\frac{2}{3}$
• C. $1$
• D. None of these

Answer 1

Answer: (D)

None of these

By given equation $y^2=4ax$ is a right handed parabola

Let $P=(a{t}^2,2at)$ since it lies on the parabola $N$ is th efoot of perpendicular from $P$ on its aixs

$\therefore$ coordinate of $N=(a{t}^2,0)$

Let $l$ be the line parallel to the x-axis which bisect $NP$ at $B$

$\therefore$ $B=(a{t}^2,at)$

Then coordinate of $Q$ in the curve $=(\frac{y^2}{4a},at)$

$=(\frac{{\left(at\right)}^2}{4a},at)$

$=(\frac{a{t}^2}{4},at)$

Let coordinates of $T$ be $(0,c)$

(since $A$ is the vertex i.e $(0,0)$ the tangent at the vertex is the y-axis)

Now since the line $l$ and the x-axis are parallel and $NT$ is a linetraversing through $l$ and x-axis the angles $\angle TQD=\angle BQN=\theta$ (vertically opposite angles are equal)

In $△TQD$, $\tan \theta =\frac{c-at}{\frac{a{t}^2}{4}}$

and the $△BQN$, $\tan \theta =\frac{at}{a{t}^2-\frac{a{t}^2}{4}}$

$\Rightarrow$ $\tan \theta =\frac{c-at}{\frac{a{t}^2}{4}}=\frac{at}{\frac{3a{t}^2}{4}}$

$\Rightarrow$ $c-at=\frac{at}{3}$

$\Rightarrow$ 3c-3at=at$

$\Rightarrow$ 4at=3c$

$\Rightarrow$ $c=\frac{4}{3}at$

Coordinate of $T=(0,\frac{4}{3}at)$

Now it is given that $AT=kNP$

$\frac{4}{3}at=k\left(2at\right)$

$\Rightarrow$ $k=\frac{4}{3}at.\frac{1}{2at}$

$\Rightarrow$ $k=\frac{2}{3}$