Question

$a{x}^2+2hxy+b{y}^2=0$ represents a pair of straight lines through origin & angle between them is given by

$\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}$. If the lines are perpendicular then $a+b=0$ and the equation of bisectors is given by $\frac{x^2-y^2}{a-b}=\frac{xy}{h}$

The general equation of second degree given by

$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ represent a pair of straight lines if $△=0$ or

$\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|=0$ or $abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$

On the basis of above information answer the following question

The equation $8x^2+8xy+2y^2+26x+13y+15=0$ represents a pair of parallel straight lines $2x+y+c_1=0$ & $2x+y+c_2=0$ then $c_1$ & $c_2$ are

• A. $-5,-\frac{3}{2}$
• B. $5,\frac{3}{2}$
• C. $-5,\frac{3}{2}$
• D. $5,-\frac{3}{2}$

Answer 1

Answer: (B)

$5,\frac{3}{2}$

Given equation:

$8x^2+8xy+2y^2+26x+13y+15=0$

can be written as $2{(2x+y)}^2+13(2x+y)+15=0$

or $2t^2+13t+15=0$ (where $t=2x+y$)

$\Rightarrow (t+5)(2t+3)=0$

$\Rightarrow t+5=0and2t+3=0$

$\Rightarrow 2x+y+5=0and2(2x+y)+3=0$

$\Rightarrow 2x+y+5=0and2x+y+\frac{3}{2}=0$

$\Rightarrow 2x+y+c_1=0and2x+y+c_2=0$

$\Rightarrow c_1=5then{c}_1=\frac{3}{2}$

if $c_1=\frac{3}{2}then{c}_1=5$

$\therefore c_1$ & $c_2$ are $5,\frac{3}{2}or\frac{3}{2},5$