Question

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 :

• A. $L_1 > L_2$
• B. $L_1 = L_2$
• C. $L_1 < L_2$
• D. $\frac{L_{1}}{L_{2}} = \sqrt{2}$

Answer 1

Answer: (C)

$L_1 < L_2$

Answer 2

$x^{2} + y^{2} = 9 \,\&\, y^{2} = 8x$
$L_{2} =$ L.R. of $y^{2} = 8x \Rightarrow L_{2} = 8$
Solve $x^{2} + 8x = 9 \Rightarrow x = 1,-9$
$x = -9 reject$
$\therefore y^{2} = 8 \,x$ so $y^{2} = 8$
$y = \pm \sqrt{8}$
Point of intersection are $\left(1, \sqrt{8}\right) \left(1, - \sqrt{8}\right)$
So $L_{1} = 2\sqrt{8}$
$\frac{L_{1}}{L_{2}} = \frac{2\sqrt{8}}{8} = \frac{2}{\sqrt{8}} = \frac{1}{\sqrt{2}} < 1$
$L_{1} < L_{2}$