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Q.
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
Conic Sections
Solution:
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 $
$\because(1,-1) is on it \therefore k =2$
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) $
