Question

The gradient of one of the lines of ${{x}^{2}}+hxy+2{{y}^{2}}=0$ is twice that of the other, then $h=$
A. $\pm 3$
B. $\pm \dfrac{3}{2}$
C. $\pm 2$
D. $\pm 1$

Answer 1

Hint: In this question, we are to find the value of the given variable. For this, we use the equation formed by the gradients of the given two lines. So, 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
${{x}^{2}}+hxy+2{{y}^{2}}=0$
By comparing with the equation $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$, we get
$a=1;h=\dfrac{h}{2};b=2$
Here, for calculating the required variable, consider $\dfrac{h}{2}$ as its value.
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
\[{{m}_{1}}+{{m}_{2}}=\dfrac{-2h}{b}\text{=}\dfrac{-h}{2}\text{ }...(2)\]
And the product of these two gradients is
${{m}_{1}}{{m}_{2}}=\dfrac{a}{b}=\dfrac{1}{2}\text{ }...(3)$
Substituting (1) in (2), we get
$\begin{align}
  & 2{{m}_{2}}+{{m}_{2}}=\dfrac{-2h}{b} \\
 & \Rightarrow 3{{m}_{2}}=\dfrac{-2h}{b} \\
 & \Rightarrow {{m}_{2}}=\dfrac{-2h}{3b}\text{=}\dfrac{-h}{6}\text{ }...(4) \\
\end{align}$
Substituting (1) in (3) and on simplifying, we get
$\begin{align}
  & {{m}_{1}}{{m}_{2}}=\dfrac{1}{2} \\
 & \Rightarrow 2{{m}_{2}}^{2}=\dfrac{1}{2} \\
 & \Rightarrow 2{{\left( \dfrac{-h}{6} \right)}^{2}}=\dfrac{1}{2} \\
 & \Rightarrow \dfrac{2{{h}^{2}}}{36}=\dfrac{1}{2} \\
 & \Rightarrow {{h}^{2}}=9 \\
 & \therefore h=\pm 3 \\
\end{align}$
Thus, the value is $h=\pm 3$.



Option ‘C’ is correct



Note: Here to find the required variable value. For this, we use the equation that relates the coefficients of the equation of pair of straight lines with their gradients. So, we need to use the sum and product of the gradients of these two straight lines.