Let P be a point on the parabola, y2 = 12x and N be the foot of the perpendicular drawn fr…

Question

Let P be a point on the parabola, y2
 = 12x and
N be the foot of the perpendicular drawn from
P on the axis of the parabola. A line is now
drawn through the mid-point M of PN, parallel
to its axis which meets the parabola at Q. If the
y-intercept of the line NQ is $${4 \over 3}$$,
 then :

Accepted Answer

Let P = (3t2
 , 6t) ; N = (3t2
, 0)

M = (3t2
, 3t)

Let point Q = (h, 3t) which lies on the parabola y2 = 12x

$$ \therefore $$ 9t2 = 12h

$$ \Rightarrow $$ h = $${{3{t^2}} \over 4}$$

Equation of NQ

y - 0 = $${{3t} \over {{{3{t^2}} \over 4} - 3{t^2}}}\left( {x - 3{t^2}} \right)$$

$$ \Rightarrow $$ y = $${{ - 4} \over {3t}}\left( {x - 3{t^2}} \right)$$

For y-intercept of NQ, x = 0

$$ \therefore $$ y = 4t

Given y-intercept of the line NQ is $${4 \over 3}$$

$$ \therefore $$ 4t = $${4 \over 3}$$

$$ \Rightarrow $$ t = $${1 \over 3}$$

$$ \therefore $$ MQ = $${{9{t^2}} \over 4}$$ = $${1 \over 4}$$

PN = 6t = 2

Solutions · 1