JEE MAIN - Mathematics (2020 - 3rd September Morning Slot - No. 13)
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 :
MQ = $${1 \over 3}$$
PN = 4
PN = 3
MQ = $${1 \over 4}$$
Explanation
_3rd_September_Morning_Slot_en_13_1.png)
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
Comments (0)
