Question

The values of $h$ for which the equation $3x^2+2hxy-3y^2-40x+30y-75=0$ represents a pair of straight lines, are

• A. $4,4$
• B. $4,6$
• C. $4,-4$
• D. $0,4$

Answer 1

The correct option is A

$4,4$

Explanation for the correct option:

Given: $3x^2+2hxy-3y^2-40x+30y-75=0$

The general equation of the curve is $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$

By comparing the given equation with the general equation we get

$\mathrm{a}=3,\mathrm{b}=-3,\mathrm{g}=-20,\mathrm{f}=15,\mathrm{c}-75,\mathrm{h}=\mathrm{h}$

The condition for the straight line is $\mathit{a}\mathit{b}\mathit{c}\mathbf{+}\mathbf{2}\mathit{f}\mathit{g}\mathit{h}\mathbf{–}\mathit{a}{\mathit{f}}^{\mathbf{2}}\mathbf{–}\mathit{b}{\mathit{g}}^{\mathbf{2}}\mathbf{–}\mathit{c}{\mathit{h}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$

Substitute the values in straight line condition

$\begin{array}{crcc}\Rightarrow & 3·\left(-3\right)·(-75)+2·15·(-20)·h-3·{\left(15\right)}^2-(-3)·{\left(-20\right)}^2-(-75)h^2& =& 0\\ \Rightarrow & 675-600h-675+1200+75h^2& =& 0\\ \Rightarrow & 75h^2-600h+1200& =& 0\end{array}$

Divide the above equation by $75$

$\begin{array}{crc}\Rightarrow & h^2-8h+16& =0\\ \Rightarrow & {(h-4)}^2& =0\\ \Rightarrow & h& =4\\ \Rightarrow & h& =4\end{array}$

Hence option A is the correct answer.