Question

If the pair of lines $3x^2-5xy+p{y}^2=0$ and $6x^2-xy-5y^2=0$ have one line in common, then $p=$

• A. $2,\frac{25}{4}$
• B. $-2,\frac{25}{4}$
• C. $-2,\frac{-25}{4}$
• D. $2,\frac{-25}{4}$

Answer 1

Answer: (D)

$2,\frac{-25}{4}$

$6x^2-xy-5y^2=0$

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

$6t^2-t-5=0$

$\Rightarrow t=\frac{1\pm \sqrt{1+120}}{12}$

$\Rightarrow t=\frac{1\pm 11}{12}$

$\Rightarrow t=1,\frac{-5}{6}$

$y=x$ or $y=\frac{-6}{5}x$

$\therefore 3x^2-5xy+p{y}^2=0$

$t=\frac{x}{y},$ $3t^2-5t+p=0$

If, $t=1,p=2$

And If, $t=\frac{-5}{6},p=5\times \left(\frac{-5}{6}\right)-3\times \frac{25}{36}$

$=\frac{-25\times 6}{36}-\frac{75}{36}$

$=\frac{-225}{36}$

$p=\frac{-25}{4}$

So, $p=2$ or $\frac{-25}{4}$ are two values.