Exam

JEE Main

Year

2014

Date

11 April

Mode

Online

Subject

Mathematics

Chapter

Circle

JEE Main 2014 · 11 April · Online · Mathematics · Circle

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

Let $\mathrm{L}_1$ 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:

Step-by-step solution

We have\begin{aligned}&x^2+(8 x)=9 \\ &x^2+9 x-x-9=0 \\ &x(x+9)-1(x+9)=0 \\ &(x+9)(x-1)=0 \\ &x=-9,1 \\ &\text{ for }x=1, y=\pm2 \sqrt{2 x}=\pm2 \sqrt{2}\\ &\mathrm{~L}}_{1}=\text{ Length of } \mathrm{AB}\\ &\quad=\sqrt{(2 \sqrt{2}+2 \sqrt{2})^2+(1-1)^2}=4 \sqrt{2}\\ &\mathrm{~L}}_{2}=\text{ Length of latus rectum }\\ &=4 a=4 \times2=8 \\ &\mathrm{~L}}_{1} < \mathrm{L}}_{2} \end{aligned}