Question

If the slope of one of the lines represented by the equation $a{x}^2+2hxy+b{y}^2=0$ be $\lambda$ times that of the other, then

• A. $4\lambda h=ab(1+\lambda )$
• B. $\lambda h=ab(1+\lambda {)}^2$
• C. $4\lambda h^2=ab(1+\lambda {)}^2$
• D. None of these.

Answer 1

The correct option is C

$4\lambda h^2=ab(1+\lambda {)}^2$

Explanation of correct option.

Step 1: Use the relation between slopes and coefficient of a pair of lines represented by a second-degree equation

Let the slope of one of the line represented by the equation $a{x}^2+2hxy+b{y}^2=0$ be $m$, then the slope of another line will be $\lambda m$

We know that for the pairs of the equation represented by $a{x}^2+2hxy+b{y}^2=0$,

The product of the slopes of two lines $=\frac{a}{b}$

The sum of the slopes of the two lines $=-\frac{2h}{b}$

But the slopes of two lines are $m$ and $\lambda m$

$\therefore m\left(\lambda m\right)=\frac{a}{b}....\left(1\right)$ and $m+\lambda m=-\frac{2h}{b}....\left(2\right)$

Step 2: Eliminate $m$ from both equations $\left(1\right)$ and $\left(2\right)$

From $\left(2\right)$, we have

$$\begin{gathered} \Rightarrow m+\lambda m=-\frac{2h}{b} \\ \Rightarrow m(1+\lambda )=-\frac{2h}{b} \\ \Rightarrow m=-\frac{2h}{b(1+\lambda )}\dots \text{(3)} \end{gathered}$$

From $\left(1\right)$, we have

$$\begin{gathered} m\left(\lambda m\right)=\frac{a}{b} \\ \Rightarrow \lambda m^2=\frac{a}{b} \\ \Rightarrow m^2=\frac{a}{b\lambda }....\left(4\right) \end{gathered}$$

Squaring $\left(3\right)$ and equating to $\left(4\right)$, we get

$\frac{4h^2}{b^2(1+\lambda {)}^2}=\frac{a}{b\lambda }$

$\Rightarrow 4\lambda h^2=ab(1+\lambda {)}^2$

Hence, Option (C) is the correct answer.