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GRE Subject Test: Math : Conic Sections

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Example Questions

Example Question #1 : Circles

What is the equation of a circle with center at (4,-3) and a radius of ?

Possible Answers:

(x+4)^2+(y-3)^2=49

(x+4)^2+(y+3)^2=49

(x-4)^2+(y+3)^2=49

(x-4)^2+(y-3)^2=49

Correct answer:

(x-4)^2+(y+3)^2=49

Explanation:

Step 1: Recall the general equation for a circle (if the vertex is not at (0,0) :

(x-h)^2+(y-k)^2=r^2 , where center= (h,k)

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= (4,-3) , h=4 , k=-3

Step 4: Plug in h,k,r into the equation of a circle:

(x-4)^2+(y-(-3))^2=7^2
Simplify:

(x-4)^2+(y+3)^2=49

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Example Question #1 : Circles

What is the vertex of the equation of a circle: x^2+y^2=9

Possible Answers:

(-3,0)

(0,3)

(0,0)

(3,0)

Correct answer:

(0,0)

Explanation:

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 (0,0) .

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Example Question #2 : Conic Sections

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

Foci: (-5,4) and (7,4)

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

Possible Answers:

$\small \small \small \small \small \frac{(x-4)^{2}}{20} - \frac{(y-1)^{2}}{36}=1$

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

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

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

$\small \small \small \small \frac{(x-1)^{2}}{16} - \frac{(y-4)^{2}}{36}=1$

Correct answer:

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

Explanation:

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 = c/a

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

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 .

12=2c

$\small c= 6$

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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Example Question #11 : Conic Sections

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

Foci: (1, 8) and (9, 8)

Eccentricity:

Possible Answers:

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

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

$\small \small \small \small \small \small \small \frac{(x-5)^{2}}{4} - \frac{(y-8)^{2}}{16}=1$

$\small \small \small \small \small \small \small \small \frac{(x-8)^{2}}{12} - \frac{(y-5)^{2}}{16}=1$

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

Correct answer:

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

Explanation:

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 = c/a

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

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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Example Question #2 : Solve A System Of Quadratic Equations

Find the coordinate of intersection, if possible: y=x^2+4 and y=-x^2+4 .

Possible Answers:

$(0,\pm4)$

(-4,0)

$\textup{There is no intersection.}$

(4,0)

(0,4)

Correct answer:

(0,4)

Explanation:

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

-x^2+4=x^2+4

2x^2 =0

x=0

Substitute x=0 into either parabola.

y=0^2+4 = 4

The coordinate of intersection is (0,4) .

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Example Question #3 : Solve A System Of Quadratic Equations

Find the intersection(s) of the two parabolas: y=2x^2+3x+4 , y=-4x^2+4

Possible Answers:

$(0,4) \textup{ and }\left(-\frac{1}{2},3\right)$

$(0,4) \textup{ and }\left(\frac{1}{2},3\right)$

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

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

(0,4)

Correct answer:

$(0,4) \textup{ and }\left(-\frac{1}{2},3\right)$

Explanation:

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

2x^2+3x+4=-4x^2+4

6x^2+3x=0

3x(2x+1)=0

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

Substitute both values of into either parabola and determine .

y=-4(0)^2+4= 4

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

The coordinates of intersection are:

(0,4) and $\left(-\frac{1}{2},3\right)$

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Example Question #1 : Solve A System Of Quadratic Equations

Find the points of intersection:

y = x^2 + 40x - 6 ; y = -x^2 + 19 x + 5

Possible Answers:

(-11, -567) ; (-0.5, -25.75)

(11, 555) ; (0.5 , 14.25)

(-11, -567); (0.5, 9.05)

(-0.5, -25.75) ; (-11, -325)

(-11, -325); (0.5, 14.25)

Correct answer:

(-11, -325); (0.5, 14.25)

Explanation:

To solve, set both equations equal to each other:

x^2 + 40x - 6 = -x^2 + 19x + 5

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

x^2 + x^2 +40x - 19x - 6 - 5 = 0

This simplifies to

2x^2 + 21 x - 11 = 0

Solving by factoring or the quadratic formula gives the solutions and 0.5 .

Plugging each into either original equation gives us:

(-11)^2 + 40(-11 ) - 6 = 121 - 440 - 6 = -325

(0.5) ^ 2 + 40 (0.5) - 6 = 14.25

Our coordinate pairs are (-11, -325) and (0.5, 14.25 ) .

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GRE Subject Test: Math : Conic Sections

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #1 : Circles

Example Question #1 : Circles

Example Question #2 : Conic Sections

Example Question #11 : Conic Sections

Example Question #2 : Solve A System Of Quadratic Equations

Example Question #3 : Solve A System Of Quadratic Equations

Example Question #1 : Solve A System Of Quadratic Equations

All GRE Subject Test: Math Resources

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