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

If the lines joining origin to the points of intersection of the line $x+y=1$ with the curve $x^2+y^2+x-2y-m=0$ are perpendicular to each other, then value of $m$ is

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $1$
• D. $-1$

Answer 1

The correct option is A $\frac{1}{2}$

As the line $x+y=1$ intersect the curve,$x^2+y^2+x-2y-m=0$

$\Rightarrow x^2+y^2+(x-2y)\left(1\right)-m{\left(1\right)}^2=0$

$\Rightarrow x^2+y^2+(x-2y)(x+y)-m{\left(1\right)}^2=0$

$\Rightarrow x^2+y^2+(x-2y)(x+y)-m{(x+y)}^2=0$

$\Rightarrow x^2+y^2+(x^2-xy-2y^2)-m(x^2+y^2+2xy)=0$

$\Rightarrow x^2(2-m)+y^2(1-2-m)+xy(1-2m)=0$ ......(*)

Now pair of straight lines given bye (*) will be $\perp$er if

$2-m+1-2-m=0$ (Using $a+b=0$)

$\Rightarrow m=\frac{1}{2}$