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BITSAT 2024Paper 1

Exam: BITSAT
Year: 2024
Paper: Paper 1
Subject: Mathematics
Chapter: Parabola

BITSAT 2024 · Paper 1 · Mathematics · Parabola

Let L_1 be the length of the common chord of the curves x^2+y^2=9...

Let

L_1

be the length of the common chord of the curves

x^2+y^2=9

and

y^2=8 x

, and

L_2

be the length of the latus rectum of

y^2=8 x

, then:

Let L_1 be the length of the common chord of the curves x^2+y^2=9 and y^2=8 x , and L_2 be the length of the latus

Options

A $\mathrm{L}_1>\mathrm{L}_2$
B $\mathrm{L}_1=\mathrm{L}_2$
C $\mathrm{L}_1 < \mathrm{L}_2$
D $\frac{\mathrm{L}_1}{\mathrm{ L}_2}=\sqrt{2}$

Correct answer

C. $\mathrm{L}_1 < \mathrm{L}_2$

Step-by-step solution

Let L_1 be the length of the common chord of the curves x^2+y^2=9 and y^2=8 x , and L_2 be the length of the latus

We have :

x^2+(8 x)=9\Rightarrow x^2+9 x-x-9=0\Rightarrow x(x+9)-1(x+9)=0\Rightarrow(x+9)(x-1)=0 \quad \Rightarrow x=-9,1

for

x=1, y= \pm 2 \sqrt{2 x}= \pm 2 \sqrt{2}\sqrt{(2 \sqrt{2}+2 \sqrt{2})^2+(1-1)^2}=4 \sqrt{2}\mathrm{L}_n=

Length of latus rectum

=4 a=4 \times 2=8\mathrm{L}_1 < \mathrm{L}_2

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