Question

If $\frac{x^2}{a}+\frac{y^2}{b}+\frac{2xy}{h}=0$ represents a pair of straight lines and slope of one line is twice the other, then $ab:h^2$ is:

• A. $9:8$
• B. $8:9$
• C. $1:2$
• D. $2:1$

Answer 1

The correct option is A

$9:8$

Explanation for the correct option:

The given equation of a pair of pair of straight lines is $\frac{x^2}{a}+\frac{y^2}{b}+\frac{2xy}{h}=0..........\left(1\right)$

Let, the slope of the first line be $m$, then the slope of the other line will be $2m$.

So, the equation of one line will be $y-mx=0$ and of other will be $y-2mx=0$.

So, the equation of the pair of pair of straight lines will be,

$\begin{array}{rcl}\left(y-mx\right)\left(y-2mx\right)& =& 0\\ & \Rightarrow & 2m^2{x}^2-3mxy+y^2=0............\left(2\right)\end{array}$

By comparing $\left(1\right)\mathrm{and}\left(2\right)$, we get

$a=\frac{1}{2m^2},b=1,\mathrm{and}h=-\frac{2}{3m}$

So,

$\begin{array}{rcl}\frac{ab}{h^2}& =& \frac{{\displaystyle \frac{1}{2m^2}}\times 1}{{\left(-{\displaystyle \frac{2}{3m}}\right)}^2}\\ & =& \frac{9}{8}\end{array}$

Hence, option A is correct.