Question

The equation of the parabola whose focus lies at the intersection point of the lines $x+y=3$ and $x-y=1$ and directrix is $x-y+5=0$

• A. $x^2+y^2+2xy+18x+6y-25=0$
• B. $x^2+y^2+2xy+18x+6y+25=0$
• C. $x^2+y^2+2xy-18x+6y-15=0$
• D. $x^2+y^2+2xy-18x-6y+15=0$

Answer 1

The correct option is C $x^2+y^2+2xy-18x+6y-15=0$

Since, focus lies at the point of intersection of the lines $x+y=3$ and $x-y=1$
solving the equations of these lines, we get
$x=2,y=1$
$\Rightarrow$ focus $\equiv (2,1)$

Let, $P(x,y)$ be any point on the parabola. then we know that,
Distance of $P$ from focus $=$ distance of $P$ from directrix
$\therefore \sqrt{(x-2{)}^2+(y-1{)}^2}=\frac{|1.x+(-1).y+5|}{\sqrt{1+1}}$
Squaring both sides,
$(x-2{)}^2+(y-1{)}^2=\frac{(x-y+5{)}^2}{2}$
$\Rightarrow 2[x^2-4x+4+y^2-2y+1]=x^2+y^2+25-2xy+10x-10y$
$\Rightarrow x^2+y^2+2xy-18x+6y-15=0$