Question

A line bisecting the ordinate $PN$ of a point $P(a{t}^2,2at),t>0$, on the parabola $y^2=4ax$ is drawn parallel the axis to meet the curve at $Q$. if $NQ$ meets the tangent at the vertex at the point $T$, then the coordinates of $T$ are

• A. $(0,\frac{4}{3}at)$
• B. $(0,2at)$
• C. $(\frac{1}{4}a{t}^2,at)$
• D. $(0,at)$

Answer 1

The correct option is A $(0,\frac{4}{3}at)$

Equation of the line parallel to the axis and bisecting the ordinate $PN$ of the point $P(a{t}^2,2at)$ is $y=at$ which meets the parabola $y^2=4ax$ at the point $Q(\frac{1}{4}a{t}^2,at)$

Coordinates of $N$ are $(a{t}^2,0)$.

Equation of $NQ$ is $y=\frac{0-at}{a{t}^2-\frac{1}{4}a{t}^2}(x-a{t}^2)$

which meets the tangent at the vertex, $x=0$, at the point $y=\frac{4}{3}at$