Conic Sections - GRE Subject Test: Math

Card 0 of 7

Question

What is the equation of a circle with center at and a radius of ?

Answer

Step 1: Recall the general equation for a circle (if the vertex is not at :

, where center=

Step 2: Recall the shift of the graph..

If the value of is positive, it will be shown as a negative shift in the equation.
If the value of is negative, it will be shown as a positive shift in the equation.
If the value of is positive, it will be shown as a negative shift in the equation.
If the value of is negative, it will be shown as a positive shift in the equation.

Step 3: Look at the center given in the problem and find the rule(s) in step 2 that will apply:

Center=, ,

Step 4: Plug in into the equation of a circle:

Simplify:

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Question

What is the vertex of the equation of a circle:

Answer

Step 1: There are no numbers next to and , so their is no movement of the vertex..

Step 2: Recall the vertex of a circle that does not move...

The vertex of this circle is .

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Question

Using the information below, determine the equation of the hyperbola.

Foci: and

Eccentricity: $\frac{3}{2}$

Answer

General Information for Hyperbola:

Equation for horizontal transverse hyperbola:

$\small \frac{(x-h)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}}=1$

Distance between foci =

Distance between vertices =

Eccentricity =

$\small a^{2}+b^{2}=c^{2}$

Center: (h, k)

First determine the value of c. Since we know the distance between the two foci is 12, we can set that equal to .

$\small c= 6$

Next, use the eccentricity equation and the value of the eccentricity provided in the question to determine the value of a.

Eccentricity = $\small \small c/a=3/2$

$\small \frac{c}{a}=\frac{3}{2}=\frac{6}{a}$

$\small 12=3a$

$\small a=4$

$\small \small a^{2}=16$

Determine the value of $\small b^{2}$

$\small a^{2}+b^{2}=c^{2}$

$\small \small 4^{2}+b^{2}=6^{2}$

$\small \small b^{2}=20$

Determine the center point to identify the values of h and k. Since the y coordinate of the foci are 4, the center point will be on the same line. Hence, $\small k = 4$ .

Since center point is equal distance from both foci, and we know that the distance between the foci is 12, we can conclude that $\small h=1$

Center point: $\small \small (h, k)= (1,4)$

Thus, the equation of the hyperbola is:

$\small \small \frac{(x-1)^{2}}{16} - \frac{(y-4)^{2}}{20}=1$

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Question

Using the information below, determine the equation of the hyperbola.

Foci: and

Eccentricity:

Answer

General Information for Hyperbola:

Equation for horizontal transverse hyperbola:

$\small \frac{(x-h)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}}=1$

Distance between foci =

Distance between vertices =

Eccentricity =

$\small a^{2}+b^{2}=c^{2}$

Center: (h, k)

First determine the value of c. Since we know the distance between the two foci is 8, we can set that equal to .

$\small 8=2c$

$\small \small c= 4$

Next, use the eccentricity equation and the value of the eccentricity provided in the question to determine the value of a.

Eccentricity = $\small \small \small \small c/a=2$

$\small \small \frac{c}{a}=\frac{2}{1}=\frac{4}{a}$

$\small \small 4=2a$

$\small \small a=2$

$\small \small \small a^{2}=4$

Determine the value of $\small b^{2}$

$\small a^{2}+b^{2}=c^{2}$

$\small \small \small 2^{2}+b^{2}=4^{2}$

$\small \small \small b^{2}=12$

Determine the center point to identify the values of h and k. Since the y coordinate of the foci are 8, the center point will be on the same line. Hence, $\small \small k = 8$ .

Since center point is equal distance from both foci, and we know that the distance between the foci is 8, we can conclude that $\small \small h=5$

Center point: $\small \small \small (h, k)= (5, 8 )$

Thus, the equation of the hyperbola is:

$\small \small \small \small \small \small \frac{(x-5)^{2}}{4} - \frac{(y-8)^{2}}{12}=1$

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Question

Find the coordinate of intersection, if possible: and .

Answer

To solve for x and y, set both equations equal to each other and solve for x.

Substitute into either parabola.

The coordinate of intersection is .

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Question

Find the intersection(s) of the two parabolas: ,

Answer

Set both parabolas equal to each other and solve for x.

$x=0,-\frac{1}{2}$

Substitute both values of into either parabola and determine .

$y=-4\left(-\frac{1}{2}\right)^2+4= -4\left(\frac{1}{4}\right)+4= -1+4=3$

The coordinates of intersection are:

and $\left(-\frac{1}{2},3\right)$

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Question

Find the points of intersection:

;

Answer

To solve, set both equations equal to each other:

To solve as a quadratic, combine like terms by adding/subtracting all three terms from the right side to the left side:

This simplifies to

Solving by factoring or the quadratic formula gives the solutions and .

Plugging each into either original equation gives us:

Our coordinate pairs are and .

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