Question

If the pair of lines $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ intersect on the $y$-axis, then

• A. $2fgh=b{g}^2+c{h}^2$
• B. $b{g}^2\ne c{h}^2$
• C. $abc=2fgh$
• D. None of these

Answer 1

The correct option is A

$2fgh=b{g}^2+c{h}^2$

Explanation for correct option:

Given $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ ……$\left(i\right)$

The pair of lines intersect $y$ axis, $x=0$
Put $x=0$ in $\left(i\right)$

$\Rightarrow b{y}^2+2fy+c=0$

For unique point of intersection the root of the equation must be same

We know that for a quadratic equation $a{x}^2+bx+c=0$ will have equal root if $b^2-4ac=0$

Now comparing with the standard form we have,

$$\begin{gathered} \therefore f^2-bc=0 \\ \Rightarrow a{f}^2-abc=0....\left(ii\right) \end{gathered}$$

The equation $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ represents pair of straight line if

$abc+2fgh–a{f}^2–b{g}^2–c{h}^2=0......\left(iii\right)$

Now using $\left(ii\right)$ in $\left(iii\right)$ we have

$$\begin{gathered} \Rightarrow a\left(bc–f^2\right)+2fgh–b{g}^2–c{h}^2=0 \\ \Rightarrow 2fgh–b{g}^2–c{h}^2=0 \\ \Rightarrow 2fgh=b{g}^2+c{h}^2 \end{gathered}$$

Hence option (A) $2fgh=b{g}^2+c{h}^2$, is the correct answer.