Question

The equation to the pair of lines passing through the point $(-2,3)$ and parallel to the pair of lines $x^2+4xy+y^2=0$ is

• A. $x^2-4xy+y^2-8x+2y-11=0$
• B. $x^2+4xy+y^2-8x+2y-11=0$
• C. $x^2+4xy-y^2-8x-2y-11=0$
• D. $x^2-4xy+y^2-8x-2y-11=0$

Answer 1

Answer: (B)

$x^2+4xy+y^2-8x+2y-11=0$

Equation $x^2+4xy+y^2=0$ is written as $(\frac{y}{x}{)}^2+4(\frac{y}{x})+1=0$ and the solution of above equation gives the slopes of the pair of lines.

$(\frac{y}{x}{)}^2+4(\frac{y}{x})+1=0$

$m^2+4m+1=0$, where $\frac{y}{x}=m$

Solution to above quadratic equation is the slopes $m_1=-2+\sqrt{3},m_2=-2-\sqrt{3}$

The equation of pair of straight lines parallel to the given pair of straight lines has the same slopes as $m_1$ and $m_2$. It also pass t through the point $(-2,3)$. Hence we can write equation of each of the straight lines as

$y-3=(-2+\sqrt{3})(x+2)\left(1\right)$

$y-3=(-2-\sqrt{3})(x+2)\left(2\right)$

The product of (1) and (2) gives the combined equation of pair of straight lines

$\left(\right(y-3)-(-2+\sqrt{3}\left)\right(x+2\left)\right)\left(\right(y-3)-(-2-\sqrt{3}\left)\right(x+2\left)\right)=0$

$(y-3{)}^2-(y-3\left)\right(-2-\sqrt{3}\left)\right(x+2)-(-2+\sqrt{3}\left)\right(x+2\left)\right(y-3)+(x+2{)}^2=0$

which gives,

$x^2+4xy+y^2-8x+2y-11=0$