Test Index

Parabola

Section: Mathematics

Question : 7 of 32

Marks: +1, -0

Let

E

y^{2}=8x

P=(-2,4)

Q

Q'

E

PQ

PQ'

E

F

E

Solution:

Given that

E: y^{2}=8x

P=(-2,4)

Point

P(-2,4)

lies on direction of parabola

So,

\angle QPQ' = \frac{\pi}{2}

and chord

QQ'

is a focal chord and segment

PQ

subtends right angle at the focus. So,

\angle PFQ = \frac{\pi}{2}

Slope of

Q' = \frac{2}{t_{1}+t_{2}} = 1

Slope of

PF = -1

\therefore QQ' \perp PF

PF = 4\sqrt{2}

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