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Precalculus : Conic Sections

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Example Questions

Example Question #1 : Conic Sections

What is the center and radius of the circle indicated by the equation?

(x-2)^2+y^2=36

Possible Answers:

$(-2,0),\ r=36$

$(2,0),\ r=36$

$(-2,0),\ r=6$

$(2,0),\ r=6$

Correct answer:

$(2,0),\ r=6$

Explanation:

A circle is defined by an equation in the format (x-h)^2+(y-k)^2=r^2 .

The center is indicated by the point (h,k) and the radius .

In the equation (x-2)^2+(y)^2=36=6^2 , the center is (2,0) and the radius is .

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Example Question #1 : Circles

What is the equation of the circle with center at (4, -5) and a point on the circle at (4, -2)? What is the radius?

Possible Answers:

$\left ( x-4\right )^2 + (y+5)^2 = 9$

r = 3

$\left ( x-4\right )^2 + (y-5)^2 = 9$

r = 9

$\left ( x-4\right )^2 + (y+5)^2 = 3$

r = 3

$\left ( x+4\right )^2 + (y+5)^2 = 9$

r = 3

(x+4)^2-(y+5)^2 = 3

r = 9

Correct answer:

$\left ( x-4\right )^2 + (y+5)^2 = 9$

r = 3

Explanation:

The form most often used for circles is the following general equation:

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

where (h, k) are the coordinates of the center and r is the radius. We are given the coordinates of the center as (4, -5), so h is 4 and k is -5.

(x-4)^2 +(y-(-5)))^2 = r^2

(x-4)^2 +(y+5)^2 = r^2

We still need to find the radius. We can do this by plugging in the second given point, (4, -2).

(4 - 4)^2 + (-2 + 5)^2 = r^2

0^2 + 3^2 = r^2

0 + 9 = r^2

9 = r^2

Hence the formula of the circle is

(x-4)^2 +(y+5)^2 = 9 ,

and the radius is r = 3 .

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Example Question #2 : Conic Sections

Find the center and radius of this circle:

x^2+y^2-8x+2y-19=0

Possible Answers:

c=(4,-1)

r=6

c=(8,-2)

r=6

c=(4,-1)

$r=\sqrt{19}$

c=(4,-1)

$r=\sqrt{2}$

c=(-4,1)

r=6

Correct answer:

c=(4,-1)

r=6

Explanation:

To solve this problem we need to get our function into the form:

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

where center=(h,k) and r=radius. To do this we need to complete the square with both the x an y parts of the equation

(x^2-8x+16)+(y^2+2x+1)=19+16+1

then we factor our trinomials and add our constants and we get:

(x-4)^2+(y+1)^2=36

so our center is (4,-1) and our radius is 6

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Example Question #2 : Circles

Which function will graph a circle of radius , centered at (-2,8) ?

Possible Answers:

(x-2)^2+(y-8)^2=16

(x-2)^2+(y+8)^2=16

(x+2)^2+(y-8)^2=4

(x+2)^2+(y-8)^2 = 16

(x+2)^2-(y-8)^2=16

Correct answer:

(x+2)^2+(y-8)^2 = 16

Explanation:

The general form for the function of a circle of radius r and centered at (a,b) is:

(x-a)^2+(y-b)^2=r^2 .

The circle is centered at (-2,8) , therefore we substitute in for and for . The radius is so r^2 becomes .

The final form should be:

(x+2)^2+(y-8)^2=16 .

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Example Question #2 : Conic Sections

Find the focal points of the conic below:

$27x^{^{2}}+16y^{^{2}}+324x-96y+684 = 0$

Possible Answers:

$\left ( 6,-3+\sqrt{11}\right ) , \left (6,-3-\sqrt{11} \right )$

$\left ( -6+\sqrt{11},3\right ),\left ( -6-\sqrt{11},3\right )$

$\left ( -6,3+\sqrt{11}\right ) , \left (-6,3-\sqrt{11} \right )$

$\left ( -6,14\right ) , \left (-6,-8 \right )$

$\left ( -6,3+\sqrt{13}\right ) , \left (-6,3-\sqrt{13} \right )$

Correct answer:

$\left ( -6,3+\sqrt{11}\right ) , \left (-6,3-\sqrt{11} \right )$

Explanation:

The first thing we want to do is put the conic (an ellipse because the x²and the y² terms have the same sign) into a better form i.e.

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

where (h,k) is the center of our ellipse.

We will continue by completing the square for both the x and y binomials.

First we seperate them into two trinomials:

$\left ( 27x^2+324x+0\right )+\left ( 16y^2-96y+0\right )=-684$

then we pull a 27 out of the first one and a 16 out of the second

$27\left ( x^2+12x+0\right )+16\left ( y^2-6y+0\right )=-684$

then we add the correct constant to each trinomial (being sure to add the same amount to the other side of our equation.

$27\left ( x^2+12x+36\right )+16\left ( y^2-6y+9\right )=-684+27(36)+16(9)$

then we factor our trinomials and divide by 16 and 27 to get

$\frac{\left ( x+6\right )^2}{16}+\frac{\left ( y-3\right )^2}{27}=1$

so the center of our ellipse is (-6,3) and we calculate the distance from the focal points to the center by the equation:

$c^2=\left | a^2-b^2\right |$

and we know that our ellipse is stretched in the y direction because b>a so our focal points will be c displaced from our center.

with $c=\sqrt{11}$

our focal points are $\left ( -6,3+\sqrt{11}\right ) , \left (-6,3-\sqrt{11} \right )$

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Example Question #2 : Conic Sections

Find the center of this ellipse:

$33x^2+13y^2-396x+26\sqrt{7}y+850=0$

Possible Answers:

(23,-17)

$(6,-\sqrt7)$

(-23,17)

(36,7)

$(6,\sqrt7)$

Correct answer:

$(6,-\sqrt7)$

Explanation:

To find the center of this ellipse we need to put it into a better form. We do this by rearranging our terms and completing the square for both our y and x terms.

$33(x^2-12x)+13(y^2+2\sqrt7y)=-850$

completing the square for both gives us this.

$33(x^2-12x+36)+13(y^2+2\sqrt7y+7)=429$

$33(x-6)^2+13(y+\sqrt7)^2=429$

we could divide by 429 but we have the information we need. The center of our ellipse is $(6,-\sqrt7 )$

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Example Question #4 : Pre Calculus

What is the equation of the elipse centered at the origin and passing through the point (5, 0) with major radius 5 and minor radius 3?

Possible Answers:

$\frac{x^2}{25} + \frac{y^2}{9} = 1$

$\frac{x^2}{3} + \frac{y^2}{5} = 1$

$\frac{x^2}{25} - \frac{y^2}{9} = 1$

$\frac{x^2}{5} + \frac{y^2}{3} = 1$

$\frac{x^2}{9} + \frac{y^2}{25} = 1$

Correct answer:

$\frac{x^2}{25} + \frac{y^2}{9} = 1$

Explanation:

The equation of an ellipse is

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

where a is the horizontal radius, b is the vertical radius, and (h, k) is the center of the ellipse. In this case we are told that the center is at the origin, or (0,0), so both h and k equal 0. That brings us to:

$\frac{(x-0)^2}{a^2} + \frac{(y-0)^2}{b^2} = 1$

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

We are told about the major and minor radiuses, but the problem does not specify which one is horizontal and which one vertical. However it does tell us that the ellipse passes through the point (5, 0), which is in a horizontal line with the center, (0, 0). Therefore the horizontal radius is 5.

The vertical radius must then be 3. We can now plug these in:

$\frac{x^2}{5^2} + \frac{y^2}{3^2} = 1$

$\frac{x^2}{25} + \frac{y^2}{9} = 1$

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Example Question #1 : Ellipses

An ellipse is centered at (-3, 2) and passes through the points (-3, 6) and (4, 2). Determine the equation of this eclipse.

Possible Answers:

$\frac{(x+3)^2}{16} + \frac{(y-2)^2}{49} = 1$

$\frac{(x-3)^2}{49} + \frac{(y-2)^2}{16} = 1$

$\frac{(x+3)^2}{49} + \frac{(y-2)^2}{16} = 1$

$\frac{(x-3)^2}{49} + \frac{(y+2)^2}{16} = 1$

$\frac{(x+3)^2}{7} + \frac{(y-2)^2}{4} = 1$

Correct answer:

$\frac{(x+3)^2}{49} + \frac{(y-2)^2}{16} = 1$

Explanation:

The usual form for an ellipse is

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

where (h, k) is the center of the ellipse, a is the horizontal radius, and b is the vertical radius.

Plug in the coordinate pair:

$\frac{(x-(-3))^2}{a^2} + \frac{(y-2)^2}{b^2} = 1$

$\frac{(x+3)^2}{a^2} + \frac{(y-2)^2}{b^2} = 1$

Now we have to find the horizontal radius and the vertical radius. Let's compare points; we are told the ellipse passes through the point (-3, 6), which is vertically aligned with the center. Therefore the vertical radius is 4.

Similarly, the ellipse passes through the point (4, 2), which is horizontally aligned with the center. This means the horizontal radius must be 7.

Substitute:

$\frac{(x+3)^2}{7^2} + \frac{(y-2)^2}{4^2} = 1$

$\frac{(x+3)^2}{49} + \frac{(y-2)^2}{16} = 1$

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Example Question #3 : Conic Sections

What is the shape of the graph indicated by the equation?

$\frac{x^{2}}{16}+\frac{y^2}{4}=1$

Possible Answers:

Parabola

Circle

Ellipse

Hyperbola

Correct answer:

Ellipse

Explanation:

An ellipse has an equation that can be written in the format $\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$ . The center is indicated by (h,k) , or in this case (0,0) .

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Example Question #4 : Conic Sections

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

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

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

Possible Answers:

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

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

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

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

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

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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Precalculus : Conic Sections

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Conic Sections

Example Question #1 : Circles

Example Question #2 : Conic Sections

Example Question #2 : Circles

Example Question #2 : Conic Sections

Example Question #2 : Conic Sections

Example Question #4 : Pre Calculus

Example Question #1 : Ellipses

Example Question #3 : Conic Sections

Example Question #4 : Conic Sections

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