Question

Slope of a bisector of the angle between the lines $4x^2-16xy-7y^2=0$ is

• A. $\frac{11+\sqrt{377}}{16}$
• B. $\frac{11-\sqrt{377}}{16}$
• C. $\frac{-3+2\sqrt{3}}{7}$
• D. $\frac{-3-2\sqrt{3}}{7}$

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{11+\sqrt{377}}{16}$
B $\frac{11-\sqrt{377}}{16}$

Given pair

$4x^2-16xy-7y^2=0$

On comparing given eq with $a{x}^2+2hxy+b{y}^2=0$

$a=4,b=-7,h=-8$

Eq of pair of Angle bisector

$\frac{x^2-y^2}{a-b}=\frac{xy}{h}$

$\frac{x^2-y^2}{11}=\frac{xy}{-8}$

$-8x^2+8y^2=11xy$

$8y^2-11xy-8x^2=0$

$y=\frac{-(-11x)\pm \sqrt{121x^2+256x^2}}{16}$

$y=\frac{11x\pm \sqrt{377x^2}}{16}$

$y=\frac{11x\pm \sqrt{377}x}{16}$

$y=\left(\frac{11\pm \sqrt{377}}{16}\right)x$

Comparing above eq with $y=mx+c$

$m=\left(\frac{11+\sqrt{377}}{16}\right)$ and $\left(\frac{11-\sqrt{377}}{16}\right)$