Test Index

Parabola Part 1

Section: Mathematics

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Question : 35 of 100

Marks: +1, -0

Consider the parabola with vertex

\left(\frac{1}{2}, \frac{3}{4}\right)

and the directrix

y=\frac{1}{2}

. Let P be the point where the parabola meets the line

x=-\frac{1}{2}

. If the normal to the parabola at P intersects the parabola again at the point Q, then

(PQ)^2

[1 Sep 2021 Shift 2]

$\frac{75}{8}$
$\frac{125}{16}$
$\frac{25}{2}$
$\frac{15}{2}$

Solution:

Vertex

\left(\frac{1}{2}, \frac{3}{4}\right)

Equation of directrix

y=\frac{1}{2}

Equation of parabola is

\left(x-\frac{1}{2}\right)^2=y-\frac{3}{4}

Point on parabola

P\left(-\frac{1}{2}, \frac{7}{4}\right)

Equation of normal at

P\left(-\frac{1}{2}, \frac{7}{4}\right)

is

x=2y-4

This normal cuts the parabola at

Q(2,3)

(PQ)^2=\left(2+\frac{1}{2}\right)^2+\left(3-\frac{7}{4}\right)^2=\frac{125}{16}

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