Question

The axis of parabola is along the line $x+y=0$ and the distance of the vetex and focus from the origin is $\sqrt{2}$, then equation of parabola can be

• A. $(x+y{)}^2=-16(x-y+2)$
• B. $(x+y{)}^s=16(x-y+2)$
• C. $(x+y{)}^2=-16(x-y-2)$
• D. $(x+y{)}^2-16(x-y-2)$

Answer 1

Answer: (B), (C)

Points at a distance $\sqrt{2}$ from origin an line $x+y=0$ are $(-1,1)$ and $(1,-1)$
(i) If vertex is $(-1,1)$ and focus is $(1,-1)$, then equation of tangent at vertex will be $x-y=\lambda$
$(-1,1)$ is on it
$\therefore x-y=-2$
Distance between focus and vertex is $2\sqrt{2}$.
Equation of parabola will be
${\left(\frac{x+y}{\sqrt{2}}\right)}^2\pm 4\times 2\sqrt{2}\left(\frac{x-y+2}{\sqrt{2}}\right)=0\{$ Only one sign is possible \}
but $(0,0)$ lies inside the parabola
$\therefore -$ sign must be taken
$\therefore$ required equation of parabola will be ${\left(\frac{x+y}{\sqrt{2}}\right)}^2-4\cdot 2\sqrt{2}\left(\frac{x-y+2}{\sqrt{2}}\right)=0$
$\Rightarrow (x+y{)}^2=16(x-y+2)$
(ii) If vertex is $(1,-1)$ and focus is $(-1,1)$
Then equation of tangent at vertex
$x-y=k$
$\because (1,-1)isonit\therefore k=2$
Equation of parabola will be
${\left(\frac{x+y}{\sqrt{2}}\right)}^2\pm 4\cdot 2\sqrt{2}\left(\frac{x-y-2}{\sqrt{2}}\right)=0$ (Onlyone sign is possible)
but $(0,0)$ lies inside the parabola
$\therefore$ +sign must be taken
$\therefore {\left(\frac{x+y}{\sqrt{2}}\right)}^2+4\cdot 2\sqrt{2}\left(\frac{x-y-2}{\sqrt{2}}\right)=0$
$(x+y{)}^2=-16(x-y-2)$