Test Index

Parabola

Section: Mathematics

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Question : 15 of 15

Marks: +1, -0

The line

x-2 y-3=0

cuts the parabola

y^2=4 a x

P

Q

. If the focus of this parabola is

\left(\;\frac{1}{4}, k\right)

PQ=

[AP EAPCET 18 May 2024 Shift 1]

$16 a \sqrt{5}$
$8 a \sqrt{5}$
$4 a \sqrt{5}$
$2 a \sqrt{5}$

Solution:

Focus of the parabola

y^2=42 x

is

(\alpha, 0)

Here,

a=\;\frac{1}{4}

and

K=0

So, equation of parabola is

y^2=x

As, the line

x=2 y+3

cuts the parabola

y^2\;=x

y^2\;=2 y+3

\Rightarrow y^2-2 y-3\;=0

y^2-3 y+y-3\;=0

y\;=3,-1

If

y=3

, then

x=9

and

If

y=-1

, then

x=1

The two intersecting points

P

and

Q

are (9.3) and

(1,-1)

P Q\;=\sqrt{(9-1)^2+(3+1)^2}

\;=\sqrt{64+16}=\sqrt{80}

\;=4\sqrt{5}

\;=16 a \sqrt{5}\;\;[\because a=\;\frac{1}{4}]

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