Mathematics - Parabola Question with Solution | TestHub

As distance of vertex from origin is $\sqrt{2}$ and focus is $2\sqrt{2}$. $\therefore V(1,1)$ and $F(2,2)$ [i.e. lying on $y=x$ ] where, length of latusrectum $$\begin{array}{rl}& =4a,\\ & =4\sqrt{2}\end{array}$$ $$[\because \text{ where }a=\sqrt{2}]$$ $\therefore$ By definition of parabola $$P{M}^2=(4a)(PN)$$ where, $PN$ is length of perpendicular upon $x+y-2=0$ [i.e. tangent at vertex] $$\begin{array}{ll}\Rightarrow & \frac{(x-y{)}^2}{2}=4\sqrt{2}\left(\frac{x+y-2}{\sqrt{2}}\right)\\ \therefore & (x-y{)}^2=8(x+y-2)\end{array}$$ Hence, the equation of parabola is $(x-y{)}^2=8(x+y-2)$.