Question

The sum of the squares of the slopes of the lines $x^2-xy-6y^2=0$ is

• A. $\frac{1}{36}$
• B. $\frac{5}{36}$
• C. $\frac{13}{36}$
• D. $\frac{2}{13}$

Answer 1

The correct option is C $\frac{13}{36}$

$x^2-xy-6y^2=0$

Substitute $t=\frac{x}{y}$

$t^2-t-6=0$

$\Rightarrow t=\frac{1\pm \sqrt{1+24}}{2}$

$\Rightarrow t=\frac{1\pm 5}{2}$

$\Rightarrow t=3,-2$

$\Rightarrow y=\frac{x}{3}$ & $y=\frac{-x}{2}$

Sum of squares of slopes $=(\frac{1}{3}{)}^2+(\frac{-1}{2}{)}^2$

$=\frac{1}{9}+\frac{1}{4}$

$=\frac{13}{36}$