Problem 221

1. Analysis

This problem tests the application of the parabola's equation, which is a basic question.

Solution

The focus of the parabola is $(0, \frac{1}{16})$. Therefore, the focal distance of the parabola $4x^{2}=y$ is $\frac{1}{8}$.

So, the answer is $\boxed{\frac{1}{8}}$.

2. Analysis

This problem tests the equation of the asymptotes of a hyperbola and the standard equation of a hyperbola, which is a basic question.

Solution

Let the equation of the required hyperbola be $\frac{x^{2}}{2} -y^{2}=m$. Then $\frac{4}{2}-0=m$, so $m=2$. Therefore, the required hyperbola equation is $\frac{x^{2}}{4}-\frac{y^{2}}{2}=1$.

So, the answer is $\boxed{\frac{x^{2}}{4}-\frac{y^{2}}{2}=1}$.

3. Analysis

This problem tests analogical reasoning, which is a basic question.

Solution

By analogy with the distance formula between a point $(x_{0},y_{0})$ and a line $Ax+By+C=0$ in a plane, the distance between the point $(0,1,3)$ and the plane $x+2y+3z+3=0$ is $\frac{|0\times1+2\times1+3\times3+3|}{\sqrt{1^{2}+2^{2}+3^{2}}}=\sqrt{14}$.

So, the answer is $\boxed{\sqrt{14}}$.

4. Analysis

This problem tests the concept of ellipses and the application of standard equations, which is an intermediate question.

Solution

The ellipse $5y^{2}+9x^{2}=45$ can be written as $\frac{y^{2}}{9}+\frac{x^{2}}{5}=1$. Hence, $a=3$, $b=\sqrt{5}$, and $c=2$. Since $|PF_{1}|+|PF_{2}|=2a=6$, we have $|PF_{1}|=6-|PF_{2}|$. Therefore, $|PF_{1}|+|PA|=6-|PF_{2}|+|PA|=6+(|PA|-|PF_{2}|)$. According to the triangle inequality, when point $P$ is the intersection point of the extension line of $P_{1}F_{2}$ and the ellipse, $(|PA|-|PF_{2}|)$ is the largest. At this time, the maximum value of $|PA|+|PF_{1}|$ is $M=6+\sqrt{2}$. When point $P$ is the intersection point of the extension line of $F_{2}P_{1}$ and the ellipse, $(|PA|-|PF_{2}|)$ is the smallest. At this time, the minimum value of $|PA|+|PF_{1}|$ is $N=6-\sqrt{2}$. Therefore, $M-N=2\sqrt{2}$.

So, the answer is $\boxed{2\sqrt{2}}$.