3. For the parabola $y ^ { 2 } = 2 p x ( p > 0 )$, the distance from its focus to the line $y = x + 1$ is $\sqrt { 2 }$. Then $p =$ A. 1 B. 2 C. $2 \sqrt { 2 }$ D. 4 【Answer】B 【Solution】 【Analysis】First determine the coordinates of the focus of the parabola, then use the point-to-line distance formula to find the value of $p$. 【Detailed Solution】The focus of the parabola has coordinates $\left( \frac { p } { 2 } , 0 \right)$. The distance from this point to the line $x - y + 1 = 0$ is: $\quad d = \frac { \left| \frac { p } { 2 } - 0 + 1 \right| } { \sqrt { 1 + 1 } } = \sqrt { 2 }$, Solving: $p = 2$ (we discard $p = -6$). Therefore, the answer is: B.
3. For the parabola $y ^ { 2 } = 2 p x ( p > 0 )$, the distance from its focus to the line $y = x + 1$ is $\sqrt { 2 }$. Then $p =$\\
A. 1\\
B. 2\\
C. $2 \sqrt { 2 }$\\
D. 4
【Answer】B\\
【Solution】\\
【Analysis】First determine the coordinates of the focus of the parabola, then use the point-to-line distance formula to find the value of $p$.
【Detailed Solution】The focus of the parabola has coordinates $\left( \frac { p } { 2 } , 0 \right)$. \\
The distance from this point to the line $x - y + 1 = 0$ is: $\quad d = \frac { \left| \frac { p } { 2 } - 0 + 1 \right| } { \sqrt { 1 + 1 } } = \sqrt { 2 }$, \\
Solving: $p = 2$ (we discard $p = -6$).\\
Therefore, the answer is: B.\\