Question

If the lines joining the points of intersection of the curve $4x^2+9y+18xy=1$ and the line $y=2x+c$ to the origin are equally inclined to the $4y$ axis then $c$ is

• A. $-\frac{9+\sqrt{161}}{8}$
• B. $\frac{-9+\sqrt{161}}{8}$
• C. $-1$, $-\frac{1}{2}$
• D. $\frac{1}{3},\frac{2}{3}$

Answer 1

The correct option is D $\frac{1}{3},\frac{2}{3}$

Given line

$y=2x+c$

$\frac{y-2x}{c}=1$

So from eq of pair of line

$4x^2+9y+18xy-1=0$

$4x^2+9y\left(\frac{y-2x}{c}\right)+18xy-{\left(\frac{y-2x}{c}\right)}^2=0$

$4x^2+\frac{9y^2-18xy}{c}+18xy-\frac{y^2+4x^2-4xy}{c^2}=0$

$4c^2{x}^2+9c{y}^2-18cxy+18cxy-y^2+4x^2-4xy=0$

Here both line inclined equally hence

$\tan {45}^0=2\left(\frac{\sqrt{h^2-ab}}{a+b}\right)$

$a+b=2\sqrt{h^2-ab}$

$4c^2-4+9c-1=2(9c^2-9c+1{)}^2-(4c^2-4\left)\right(9c-1)$

On solving above eq we get

$c=\frac{1}{3},\frac{2}{3}$