Question

The combined equation of two sides of a triangle is $x^2-3y^2-2xy+8y-4=0.$ The third side, which is variable always passes through the point $(-5,-1)$. If the range of values of the slope of the third line such that the origin is an interior point of the triangle is $(a,b)$, then the value of $(a+\frac{1}{b})$ is

• A. $\frac{1}{2}$
• B. $2$
• C. $0$
• D. $4$

Answer 1

The correct option is D $4$

$x^2-3y^2-2xy+8y-4=0\Rightarrow (x-3y)(x+y)+2x+2y-2x+6y-4=0\Rightarrow (x+y)(x-3y+2)-2(x-3y+2)=0\Rightarrow (x+y-2)(x-3y+2)=0$
So, the lines are $x+y-2=0$ and $x-3y+2=0$
Let the equation of the variable line be
$y+1=m(x+5)$


For origin to lie inside the required triangle $y$ intercept for the variable line should be $-\text{ive}$
$\therefore 5m-1<0\Rightarrow m<\frac{1}{5}$
$\because (-5,-1)$ also lies on $x-3y+2=0$
Now triangle can form when
$m>-1$
So, $m\in (-1,1/5)$
$\therefore a+\frac{1}{b}=4$