Question

The gradient of one of the lines of $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ is twice that of the other, then
A. ${{h}^{2}}=ab$
B. $h=a+b$
C. $8{{h}^{2}}=9ab$
D. $9{{h}^{2}}=8ab$

Answer 1

Hint: In this question, we are to find the equation formed by the gradients of the given two lines. For this, we need to use the product and sum of the slopes of a pair of straight lines. By using them, we get the required relation from the given pair of straight lines equation.



Formula Used:The combined equation of pair of straight lines is written as
$H\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}=0$
This is called a homogenous equation of second degree in $x$ and $y$
And
$S\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$
This is called a general equation of second degree in $x$ and $y$.
If ${{h}^{2}}If ${{h}^{2}}=ab$, then $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represents coincident lines.
If ${{h}^{2}}>ab$, then $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represents two real and different lines that pass through the origin.
Thus, the equation $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represents two lines. They are
$ax+hy\pm y\sqrt{{{h}^{2}}-ab}=0$
If $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represents a pair of lines, then the sum of the slopes of the lines is $\dfrac{-2h}{b}$ and the product of the slopes of the lines is $\dfrac{a}{b}$.



Complete step by step solution:Given equation is
$a{{x}^{2}}+2hxy+b{{y}^{2}}=0$
Consider the gradients of these two lines as ${{m}_{1}}$ and ${{m}_{2}}$.
It is given that, ${{m}_{1}}=2{{m}_{2}}\text{ }...(1)$
The sum of these two gradients is
${{m}_{1}}+{{m}_{2}}=\dfrac{-2h}{b}\text{ }...(2)$
And the product of these two gradients is
${{m}_{1}}{{m}_{2}}=\dfrac{a}{b}\text{ }...(3)$
Substituting (1) in (2), we get
$\begin{align}
  & 2{{m}_{2}}+{{m}_{2}}=\dfrac{-2h}{b} \\
 & \Rightarrow 3{{m}_{2}}=\dfrac{-2h}{b} \\
 & \Rightarrow {{m}_{2}}=\dfrac{-2h}{3b}\text{ }...(4) \\
\end{align}$
Substituting (1) in (3) and on simplifying, we get
$\begin{align}
  & {{m}_{1}}{{m}_{2}}=\dfrac{a}{b} \\
 & \Rightarrow 2{{m}_{2}}^{2}=\dfrac{a}{b} \\
 & \Rightarrow 2{{\left( \dfrac{-2h}{3b} \right)}^{2}}=\dfrac{a}{b} \\
 & \Rightarrow \dfrac{8{{h}^{2}}}{9{{b}^{2}}}=\dfrac{a}{b} \\
 & \Rightarrow 8{{h}^{2}}=9ab \\
\end{align}$
Thus, the required equation is $8{{h}^{2}}=9ab$.



Option ‘C’ is correct


Note: Here to find the required equation that relates the coefficients of the equation of pair of straight lines, we need to use the sum and product of the gradients of these two straight lines.