Question

If slope of one of the lines given by ax^2+2hxy+by^2=0 is four times the other then show that 16h^2=25ab​

Answer 1

Pair of Straight Lines

Solution.

The given pair of straight lines is

$\quad \mathrm{ax^{2}+2hxy+by^{2}=0}$

Let us consider

$\quad \mathrm{ax^{2}+2hxy+by^{2}=(l_{1}+m_{1}y)(l_{2}x+m_{2}y)}$

$\Rightarrow \mathrm{ax^{2}+2hxy+by^{2}=l_{1}l_{2}x^{2}+(l_{1}m_{2}+l_{2}m_{1})xy+m_{1}m_{2}y^{2}}$

Comparing both sides, we get

$\quad \mathrm{l_{1}l_{2}=a,\:m_{1}m_{2}=b,\:l_{1}m_{2}+l_{2}m_{1}=2h}$

The slope of the straight line $\mathrm{l_{1}x+m_{1}y=0}$ is

$\quad \mathrm{M=-\frac{l_{1}}{m_{1}}}$

The slope of the straight line $\mathrm{l_{2}x+m_{2}y=0}$ is

$\quad \mathrm{m=-\frac{l_{2}}{m_{2}}}$

Given that,

$\quad \mathrm{M=4m}$

$\Rightarrow \mathrm{\frac{M}{m}=\frac{4}{1}}$

$\Rightarrow \mathrm{\frac{-\frac{l_{1}}{m_{1}}}{-\frac{l_{2}}{m_{2}}}=\frac{4}{1}}$

$\Rightarrow \mathrm{\frac{l_{1}m_{2}}{l_{2}m_{1}}=\frac{4}{1}}$

Doing componendo and divedendo:

$\quad \mathrm{\frac{l_{1}m_{2}+l_{2}m_{1}}{l_{1}m_{2}-l_{2}m_{1}}=\frac{4+1}{4-1}}$

$\Rightarrow \mathrm{\frac{l_{1}m_{2}+l_{2}m_{1}}{\sqrt{(l_{1}m_{2}+l_{2}m_{1})^{2}-4l_{1}l_{2}m_{1}m_{2}}}=\frac{5}{3}}$

$\Rightarrow \mathrm{\frac{2h}{\sqrt{(2h)^{2}-4ab}}=\frac{5}{3}}$

Doing square to both sides:

$\quad \mathrm{\frac{4h^{2}}{4h^{2}-4ab}=\frac{25}{9}}$

$\Rightarrow \mathrm{\frac{h^{2}}{h^{2}-ab}=\frac{25}{9}}$

$\Rightarrow \mathrm{9h^{2}=25h^{2}-25ab}$

$\Rightarrow \color{blue}{\mathrm{16h^{2}=25ab}}$

This completes the proof.