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Parabola

Section: Mathematics

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Question : 7 of 15

Marks: +1, -0

If

L

is the normal drawn to the parabola

y^2=8 x

t=\frac{1}{\sqrt{2}}

, then the foot of the perpendicular drawn from the focus of the parabola on to the normal L is

[AP EAPCET 21 May 2025 S1]

$(3,2)$
$(5, \sqrt{2})$
$(0, \sqrt{2})$
$(3, \sqrt{2})$

Solution:

We have parabola,

y^2=8 x

Equation of normal at

(a t^2, 2 a t)

is

y+x t=2 a t+a t^3

But

t=\frac{1}{\sqrt{2}}

y+x \frac{1}{\sqrt{2}} = 4 \frac{1}{\sqrt{2}} + \frac{2}{2\sqrt{2}}

\sqrt{2} y + x = 5

It is equation of

L

.

Focus of parabola

y^2=8 x

is

(2,0)

Foot of perpendicular from focus to the line

L

is

\frac{x-2}{1} = \frac{y-0}{\sqrt{2}} = - \left( \frac{2-5}{3} \right) = 1

x=3 \text{ and } y=\sqrt{2}

\because \text{Foot of perpendicular} = (3, \sqrt{2}).

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