Test Index

Parabola Part 1

Section: Mathematics

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Question : 5 of 100

Marks: +1, -0

If

P(h,k)

be point on the parabola

x=4y^2

, which is nearest to the point

Q(0,33)

, then the distance of

P

from the directrix of the parabola

y^2=4(x+y)

[30-Jan-2023 Shift 1]

2
4
8
6

Solution:

Equation of normal

\;y=-t x+2 a t+a t^3

\;y=-t x+\;\frac{2}{16} t+\;\frac{1}{16} t^3

It passes through

(0,33)

\;33=\;\frac{t}{8}+\;\frac{t^3}{16}

\;t^3+2t-528=0

\;(t-8)(t^2+8t+66)=0

\;t=8

\;P(a t^2, 2 a t) = \left(\;\frac{1}{16} \times 64, 2 \times \;\frac{1}{16} \times 8\right) = (4,1)

Parabola :

\;y^2=4(x+y)

\;\Rightarrow y^2-4y=4x

\;\Rightarrow (y-2)^2=4(x+1)

Equation of directix :-

\;x+1=-1

\;x=-2

Distance of point

=6

Ans. : (4)

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