Test Index

Parabola Part 1

Section: Mathematics

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Question : 47 of 100

Marks: +1, -0

If the three normals drawn to the parabola,

y^{2}=2x

pass through the point

(a, 0), a \neq 0

a

must be greater than

[16 Mar 2021 Shift 1]

$\;\frac{1}{2}$
$-\;\frac{1}{2}$
$-1$
1

Solution:

Given, equation of parabola

\Rightarrow y^{2}=2x

Equation of normal of parabola,

y^{2}=4ax

is

t x+y=2at+at^{3}

Here,

4a=2

So,

a=1\frac{1}{2}

So, equation of normal

\Rightarrow t x+y=t+\frac{t^{3}}{2}

t^{3}+(2-2x)t-2y=0

As, there are three normals which are passing through

(a, 0)

, so there must be three roots of this equation. Let's say

t_{1}, t_{2}, t_{3}

.

t^{3}+(2-2a)t-2.0=0

\Rightarrow \;\; t^{3}+(2-2a)t=0

\therefore \;\; t_{1}+t_{2}+t_{3}=0

and

t_{1}t_{2}+t_{2}t_{3}+t_{3}t_{1}=2-2a

t_{1}, t_{2}, t_{3} \in R

t_{1}^{2}+t_{2}^{2}+t_{3}^{2} > 0

(t_{1}+t_{2}+t_{3})^{2}-2(t_{1}t_{2}+t_{2}t_{3}+t_{3}t_{1}) > 0

\Rightarrow 0-2(2-2a)>0 \Rightarrow 4a>4

\Rightarrow a>1

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