Question

1. 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
4
y-intercept of the line NQ is 4/5, then:
​

Answer 1

Step-by-step explanation:

ANSWER

Given equation of parabola is 

y2=4ax

Focus is (a,0)

Let P(at2,2at) be any point on the parabola.

PM is perpendicular to directrix.

Coordinates of foot of perpendicular M on directrix will be (−a,2at)

Now, PM=(at2+a)2

⇒PM=at2+a

PS=(at−a)2+4a2t2

⇒PS=at2+a

MS=4a2+4a2t2

⇒MS=2a1+t2

Since, △PMS is an equilateral triangle

PS2=MS2

⇒(at2+a)2=4a2(1+t2)

⇒t4−2t2−3=0

⇒t2=3;−1 (not possible)

⇒t2=3

So, PS=4a

Answer 2

Answer:

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