Question

The equation of parabola whose latus rectum is $2$ units, axis is $x+y-2=0$ and tangent at the vertex is $x-y+4=0$ is given by

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

Answer 1

The correct option is C $(x+y-2{)}^2=2\sqrt{2}(x-y+4)$

Given latus rectum $4a=2$

$\Rightarrow a=\frac{1}{2}$

Axis is $x+y=2$

$\Rightarrow$ (perpendicular distance from axis)${}^2=4a$ (distance from tangent)

${\left(\frac{x+y-2}{\sqrt{2}}\right)}^2=4\left(\frac{1}{2}\right)\left(\frac{x-y+4}{\sqrt{2}}\right)$

$(x+y-2{)}^2=2\sqrt{2}(x-y+4)$