Question

Through a point $A$ on the $X-axis$ a straight line is drawn parallel to $Y-axis$ so as to meet the pair of straight lines $a{x}^2+2hxy+b{y}^2=0$ in $B$ and $C$. If $AB=BC$, Then:

• A. $h^2=4ab$
• B. $8h^2=9ab$
• C. $9h^2=8ab$
• D. $9h^2=4ab$

Answer 1

The correct option is B

$8h^2=9ab$

Step 1: The coordinate of point $A,B$ and $C$

We have given, $A{x}^2+2hxy+B{y}^2=0$

Since the equation of the pair of lines does not have a constant term, we can safely assume that both lines pass through the origin.

Let the coordinates of point $A$ is $(z,0)$

Hence substituting the value of $x=z$ in the equation of the pair of lines, we get two values of $y$ corresponding to points $B$ and $C$.

Therefore by substituting $x=z$ in $b{y}^2+2hxy+a{x}^2=0$, we get

$b{y}^2+2hzy+a{z}^2=0$

The roots of $y_1$ and$y_2$are:

$$\begin{gathered} y_1=\frac{-zh+z\sqrt{h^2-ab}}{b} \\ {y}_2=\frac{-zh-z\sqrt{h^2-ab}}{b} \end{gathered}$$

Therefore,

$A=(z,0)$

$B=(z,y_1)$

$C=(z,y_2)$

Step 2: Use distance formula and find relation between $h,a$ and $b$

Applying the condition of $AB=BC$and taking the positive value of the square roots, we get;
$$\begin{gathered} \sqrt{{\left(y_1\right)}^2}=\sqrt{{\left(y_{2-}{y}_1\right)}^2} \\ \Rightarrow y_1=\left(y_2-y_1\right) \\ \Rightarrow y_2=2y_1 \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{-zh-z\sqrt{h^2-ab}}{b}=\frac{-2zh+2z\sqrt{h^2-ab}}{b} \\ \Rightarrow zh=3z\sqrt{h^2-ab} \\ \Rightarrow h=3\sqrt{h^2-ab} \\ \Rightarrow h^2=9\left(h^2-ab\right) \\ \Rightarrow 8h^2=9ab \end{gathered}$$

Hence, the correct option is B