Conics

2010006004

Level: 
C
The equation of a parabola is given by \( x^2 -8x +3y-2=0 \). Find the equation of a line, which passes through the vertex of this parabola and is parallel to the line \( 2x-5y+8=0 \).
\( -2x+5y-22 = 0 \)
\( 2x-5y-22 = 0 \)
\( 2x-5y-38 = 0 \)
\( 2x-5y+38 = 0 \)
\( -2x+5y+22 = 0 \)

9000104801

Level: 
C
Consider the hyperbola \[ xy = -1 \] and a line \(p\) parallel to one of the axes but not identical with this axis. Find the true statement.
The line \(p\) has a unique intersection with the hyperbola.
The line \(p\) has two intersections with the hyperbola.
The line \(p\) does not have any intersection with the hyperbola.
We cannot draw any conclusion on the number of intersections of the line \(p\) with the hyperbola.

9000104803

Level: 
C
Consider the hyperbola \[ \frac{x^{2}} {16} -\frac{y^{2}} {4} = 1 \] and a line \(p\) parallel to one of the axes. Find the true statement.
We cannot draw any conclusion on the number of intersections of the line \(p\) with the hyperbola.
The line \(p\) has two intersections with the hyperbola.
The line \(p\) has a unique intersection with the hyperbola.
The line \(p\) does not have any intersection with the hyperbola.

9000104805

Level: 
C
Find the slope of a line through the center of the hyperbola \[ \frac{(x - 2)^{2}} {4} -\frac{(y + 3)^{2}} {9} = 1 \] which has a unique intersection with this hyperbola.
There is no solution, the line with these properties does not exist.
\(\frac{3} {2}\)
\(-\frac{3} {2}\)
\(\frac{2} {3}\)
\(1\)
\(0\)

9000104809

Level: 
C
Among the following lines (which all pass through the point \([-1;3]\)) identify a line which is tangent to the following hyperbola. \[ (x + 2)\cdot (y - 2) = 1 \]
\(k\colon \ y = -x + 2\)
\(p\colon \ y = 3\)
\(q\colon \ x = -1\)
\(r\colon \ y = x + 4\)
None of the given answers is correct.

9000106901

Level: 
C
A body is thrown at the initial angle \(\alpha = 45^{\circ }\) and the initial velocity \(v_{0} = 10\, \mathrm{m}/\mathrm{s}\). The trajectory of the body is a parabola. Find the equation of this parabola. Hint: The coordinates of the moving body as functions of time are \[ \begin{aligned}x& = v_{0}t\cdot \cos \alpha , & \\y& = v_{0}t\cdot \sin \alpha -\frac{1} {2}gt^{2}. \\ \end{aligned} \] Consider the standard acceleration due to gravity \(g = 10\, \mathrm{m}/\mathrm{s}^{2}\).
\((x - 5)^{2} = -10\cdot (y - 2.5)\)
\((x - 5)^{2} = 10\cdot (y + 2.5)\)
\(x^{2} = -10\cdot (y - 5)\)
\((x - 5)^{2} = -10\cdot (y + 2.5)\)