Question

Equation of the parabola whose axis is y = x, distance from origin to vertex is $\sqrt{2}$ and distance from origin to focus is $2\sqrt{2},$ is (Focus and vertex lie in Ist quadrant) :

• A. $(x+y{)}^2=2(x+y-2)$
• B. $(x-y{)}^2=8(x+y-2)$
• C. $(x-y{)}^2=4(x+y-2)$
• D. $(x+y{)}^2=4(x+y-2)$

Answer 1

The correct option is B

$(x-y{)}^2=8(x+y-2)$

The equation of axis of the parabolain parametric form is $\frac{x-0}{cos{45}^o}=\frac{y-0}{sin{45}^o}=\sqrt{2}forA,2\sqrt{2}forS$
$\therefore$ A is (1, 1) and S is (2, 2) and foot of directrix be z, then A is mid-point of SZ

$\therefore \frac{x+2}{2}=1,\frac{y+2}{2}=1,\therefore zis(0,0).$
Equation of directrix is y - 0 = -1 (x - 0) or x + y = 0
By definition if P(x,y) be any point on the parabola then SP = PM
$or(x-2{)}^2+(y-2{)}^2={\left[\begin{array}{c}\frac{x+y}{\sqrt{2}}\end{array}\right]}^2$
$2[x^2+y^2-4x-4y+8]=(x+y{)}^2$
or $x^2+y^2-2xy=8(x+y-2)$