The parabola $y ^ { 2 } = 2 p x ( p > 0 )$ has its focus at a distance of $\sqrt { 2 }$ from the line $y = x + 1$. Then $p =$ A. 1 B. 2 C. $2 \sqrt { 2 }$ D. 4
B The focus of the parabola is at $\left( \frac { p } { 2 } , 0 \right)$. Its distance to the line $x - y + 1 = 0$ is: $d = \frac { \left| \frac { p } { 2 } - 0 + 1 \right| } { \sqrt { 1 + 1 } } = \sqrt { 2 }$. Solving: $p = 2$ (we discard $p = -6$).
The parabola $y ^ { 2 } = 2 p x ( p > 0 )$ has its focus at a distance of $\sqrt { 2 }$ from the line $y = x + 1$. Then $p =$
A. 1
B. 2
C. $2 \sqrt { 2 }$
D. 4