Find the distance between the points where the
\(y\)-axis
intersects the following hyperbola.
\[
H\colon \frac{\left (x - 4\right )^{2}}
{10} -\frac{\left (y - 5\right )^{2}}
{15} = 1
\]
Find the distance between the points where the
\(x\)-axis
intersects the following hyperbola.
\[
H\colon \frac{\left (x - 1\right )^{2}}
{10} -\frac{\left (y - 3\right )^{2}}
{6} = 1
\]
Find the distance between the points where the
\(y\)-axis
intersects the following hyperbola.
\[
H\colon \frac{\left (x - 4\right )^{2}}
{8} -\frac{\left (y - 3\right )^{2}}
{1} = 1
\]
Find the third coordinate of the point
\(B = [2;0;?]\) using the fact that this
point is in the plane \(\alpha \)
defined by the equation
\[
\alpha \colon 2x + y - z - 5 = 0.
\]
Use the point \(B\) to
find the angle \(\varphi \)
between the plane \(\alpha \)
and the line \(AB\),
where \(A = [0;0;1]\).
Find the general equation of the plane which is perpendicular to the plane
\(\alpha \)
\[
\alpha \colon 2x + y - z - 5 = 0
\]
and contains the line \(AB\),
where \(A = [0;0;1]\)
and \(B\) is a
point in \(\alpha \)
defined by it's first two coordinates
\[
B = [2;0;?].
\]
In the following list identify a pair of planes such that the distance of planes from the plane $\alpha$ is the same as the distance between the point $A=[0;0;1]$ and the plane \(\alpha \).
\[
\alpha \colon 2x + y - z - 5 = 0
\]
\(\begin{aligned}[t] 2x + y - z +\phantom{ 1}1& = 0&
\\2x + y - z - 11& = 0
\\ \end{aligned}\)
\(\begin{aligned}[t] 2x + y - z +\phantom{ 1}1& = 0&
\\2x + y - z - 10& = 0
\\ \end{aligned}\)
\(\begin{aligned}[t] 2x + y - z +\phantom{ 1}1& = 0&
\\2x + y - z - 12& = 0
\\ \end{aligned}\)
\(\begin{aligned}[t] 2x + y - z + 1& = 0&
\\2x + y - z - 9& = 0
\\ \end{aligned}\)
Find the distance between the points where the
\(x\)-axis
intersects the following hyperbola.
\[
H\colon \frac{\left (x - 3\right )^{2}}
{20} -\frac{\left (y - 2\right )^{2}}
{5} = 1
\]
Find the distance between the points of intersection of the given hyperbola with the given straight line.
\[
H\colon \frac{\left (x - 6\right )^{2}}
{10} -\frac{\left (y - 2\right )^{2}}
{6} = 1;\quad p\colon x - 11 = 0
\]
The plane \(\alpha \)
has equation
\[
\alpha : 2x + y - z - 5 = 0.
\]
The line \(k\) passes
through the point \(A = [0;0;1]\) and
is perpendicular to \(\alpha \).
Find the intersection \(S\)
of the line \(k\) and
the plane \(\alpha \).