Question

Find the separate equations in each of the following cases:
 $11{x^2} + 8xy + {y^2} = 0$

Answer 1

Hint:
We can divide the equation throughout with ${x^2}$. Then we can substitute for \[\dfrac{y}{x}\] as the slope. Then we can solve for the slope. Then we can find the equation for each of the slopes obtained.

Complete step by step solution:
We are given the equation $11{x^2} + 8xy + {y^2} = 0$.
We can divide the equation throughout with ${x^2}$.
 $ \Rightarrow 11 + 8\dfrac{{xy}}{{{x^2}}} + \dfrac{{{y^2}}}{{{x^2}}} = 0$
On simplification, we get
 $ \Rightarrow 11 + 8\dfrac{y}{x} + {\left( {\dfrac{y}{x}} \right)^2} = 0$
We know that slope is given by $m = \dfrac{y}{x}$. On substituting this in the equation, we get
 $ \Rightarrow 11 + 8m + {m^2} = 0$
Now we have a quadratic equation. We know that the solution of quadratic equation of the form $a{x^2} + bx + c = 0$ is given by the quadratic formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
So, the value of m is given by,
 $m = \dfrac{{ - 8 \pm \sqrt {{8^2} - 4 \times 1 \times 11} }}{{2 \times 1}}$
On simplification, we get
 $ \Rightarrow m = \dfrac{{ - 8 \pm \sqrt {64 - 44} }}{2}$
On simplifying the terms in the root, we have
 $ \Rightarrow m = \dfrac{{ - 8 \pm \sqrt {20} }}{2}$
We can write $\sqrt {20} = 2\sqrt 5 $. So, we get
 $ \Rightarrow m = \dfrac{{ - 8 \pm 2\sqrt 5 }}{2}$
On cancelling the common terms, we get
 $ \Rightarrow m = - 4 \pm \sqrt 5 $
Now we have 2 values of m.
 $ \Rightarrow m = - 4 + \sqrt 5 $ and $m = - 4 - \sqrt 5 $
When $m = - 4 + \sqrt 5 $, the equation is given by,
 $ \Rightarrow y = mx$
On substituting value of m, we get
 $ \Rightarrow y = \left( { - 4 + \sqrt 5 } \right)x$
On rearranging, we get
 $ \Rightarrow y - \left( { - 4 + \sqrt 5 } \right)x = 0$
On taking the negative sign inside the bracket, we get
 $ \Rightarrow y + \left( {4 - \sqrt 5 } \right)x = 0$
When $m = - 4 - \sqrt 5 $, the equation is given by,
 $ \Rightarrow y = mx$
On substituting value of m, we get
 $ \Rightarrow y = \left( { - 4 - \sqrt 5 } \right)x$
On rearranging, we get
 $ \Rightarrow y - \left( { - 4 - \sqrt 5 } \right)x = 0$
On taking the negative sign inside the bracket, we get
 $ \Rightarrow y + \left( {4 + \sqrt 5 } \right)x = 0$

Therefore, the separate equations of the given joint lines are $y + \left( {4 - \sqrt 5 } \right)x = 0$ and $y + \left( {4 + \sqrt 5 } \right)x = 0$

Note:
If the given equation can be factored, we can find the separate equation by the method of factorising. This method cannot be used for solving this problem as it cannot be factored. We must consider both the values of m to get 2 equations. The joint equation of 2 lines is obtained by taking the product of the equation of the separate lines.