JEE Main 2014 · 11 Apr · Shift 1 · Mathematics Q81

MathematicsSingle correct+4 / -1Question 81 of 90

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

L_1>L_2

L_1=L_2

L_1<L_2

\dfrac{L_1}{L_2}=\sqrt2

Answer and worked solution

Official answer: C

$x^2+8x-9=0$ gives $x=1$ and $y=\pm2\sqrt2$ , so $L_1=4\sqrt2\approx5.66$ . The latus rectum is $L_2=8$ . So $L_1<L_2$ .

JEE Main 2014 · 11 Apr · Shift 1, Mathematics question 81