Conic Sections - Pre-Calculus

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Question

Find the equations of the asymptotes for the hyperbola with the following equation:

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Answer

For a hyperbola with its foci on the -axis, like the one given in the equation, recall the standard form of the equation:

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$ , where is the center of the hyperbola.

Start by putting the given equation into the standard form of the equation of a hyperbola.

Group the terms together and terms together.

Factor out from the terms and from the terms.

Complete the squares. Remember to add the amount amount to both sides of the equation!

Add to both sides of the equation:

Divide both sides by .

$\frac{y^2-6y+9}{54}-\frac{x^2-2x+1}{4}=1$

Factor the two terms to get the standard form of the equation of a hyperbola.

$\frac{(y-3)^2}{54}-\frac{(x-1)^2}{4}=1$

The slopes of this hyperbola are given by the following:

$m=\pm\frac{b}{a}$

For the hyperbola in question, and $b=\sqrt{54}=3\sqrt6$ .

Thus, the slopes for its asymptotes are $\pm\frac{3\sqrt6}{2}$ .

Now, plug in the center of the hyperbola into the point-slope form of the equation of alien to get the equations for the asymptotes.

The center of the hyperbola is .

The equations for the asymptotes are then:

$y-3=\pm\frac{3\sqrt6}{2}(x-1)$

$y=\pm\frac{3\sqrt6}{2}(x-1)+3$

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Question

Find the foci of the hyperbola with the following equation:

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Answer

Recall that the standard formula of a hyperbola can come in two forms:

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

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}-=1$ , where the centers of both hyperbolas are .

First, put the given equation in the standard form of the equation of a hyperbola.

Group the terms together and the terms together.

Next, factor out from the terms and from the terms.

Complete the squares. Remember to add the same amount to both sides of the equation!

Add to both sides.

Divide both sides by .

$\frac{y^2-6y+9}{16}-\frac{x^2-4x+4}{20}=1$

Factor both terms to get the standard form of the equation of a hyperbola.

$\frac{(y-3)^2}{16}-\frac{(x-2)^2}{20}=1$

When the term with is first, that means the foci will lie on a horizontal transverse axis.

When the term with is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

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

For a hyperbola with a horizontal transverse access, the foci will be located at and .

For a hyperbola with a vertical transverse access, the foci will be located at and .

For the given hypebola in the question, the transverse axis is vertical and its center is located at .

Next, find .

$c=\sqrt{16+20}=\sqrt{36}=6$

The foci are then located at and .

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Question

Find the foci of the hyperbola with the following equation:

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Answer

Recall that the standard formula of a hyperbola can come in two forms:

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

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}-=1$ , where the centers of both hyperbolas are .

First, put the given equation in the standard form of the equation of a hyperbola.

Group the terms together and the terms together.

Next, factor out from the terms and from the terms.

Now we can complete the squares. Remember to add the same amount to both sides of the equation!

Add to both sides.

Divide both sides by .

$\frac{y^2+10y+25}{27}-\frac{x^2+2x+1}{22}=1$

Factor both terms to get the standard form of the equation of a hyperbola.

$\frac{(y+5)^2}{27}-\frac{(x+1)^2}{22}=1$

When the term with is first, that means the foci will lie on a horizontal transverse axis.

When the term with is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

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

For a hyperbola with a horizontal transverse access, the foci will be located at and .

For a hyperbola with a vertical transverse access, the foci will be located at and .

For the given hypebola in the question, the transverse axis is vertical and its center is located at .

Next, find .

$c=\sqrt{22+27}=\sqrt{49}=7$

The foci are then located at and .

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Question

Find the foci of the hyperbola with the following equation:

$\frac{x^2}{27}-\frac{y^2}{9}=1$

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Answer

Recall that the standard formula of a hyperbola can come in two forms:

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

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}-=1$ , where the centers of both hyperbolas are .

When the term with is first, that means the foci will lie on a horizontal transverse axis.

When the term with is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

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

For a hyperbola with a horizontal transverse access, the foci will be located at and .

For a hyperbola with a vertical transverse access, the foci will be located at and .

For the given hypebola in the question, the transverse axis is horizontal and its center is located at .

Next, find .

$c=\sqrt{27+9}=\sqrt{36}=6$

The foci are then located at and .

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Question

Find the foci of a hyperbola with the following equation:

$\frac{x^2}{36}-\frac{y^2}{64}=1$

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Answer

Recall that the standard formula of a hyperbola can come in two forms:

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

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}-=1$ , where the centers of both hyperbolas are .

When the term with is first, that means the foci will lie on a horizontal transverse axis.

When the term with is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

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

For a hyperbola with a horizontal transverse access, the foci will be located at and .

For a hyperbola with a vertical transverse access, the foci will be located at and .

For the given hypebola in the question, the transverse axis is horizontal and its center is located at .

Next, find .

$c=\sqrt{64+36}=\sqrt{100}=10$

The foci are then located at and .

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Question

Which is the equation for a parabola that opens down?

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Answer

The answer is because it is the only degree-2 polynomial with a negative leading coefficient.

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Question

Which of the following could NOT be a possible number of solutions of a system of quadratic and linear equations?

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Answer

Recall that the solution of a system of equations is given by the intersection points of the graphs. Thus, this question is really asking how many times a parabola and a line can intersect. Visualize a parabola and a line. For this purpose, let's say that the parabola is facing up. If a line is drawn horizontally under the vertex, then it would not intersect the parabola at all, so the system would have no solutions. If the line is tangent to, or just skims the edge, of the parabola then it would only intersect once and the system would have one solution. If the line goes straight through the parabola, then it would intersect twice. There is no other option for the orientation of the line and parabola. Thus, there cannot be three solutions.

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Question

Find the axis of symmetry and vertex of the following parabola:

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Answer

The first step of the problem is to find the axis of symmetry using the following formula:

$x_{symmetry}=-\frac{b}{2a}$

Where a and b are determined from the format for the equation of a parabola:

We can see from the equation given in the problem that a=1 and b=-3, so we can plug these values into the formula to find the axis of symmetry of our parabola:

$x_{symmetry}=-\frac{-3}{2(1)}=\frac{3}{2}$

Keep in mind that the vertex of the parabola lies directly on the axis of symmetry. That is, the x-coordinate of the axis of symmetry will be the same as that of the vertex of the parabola. Now that we know the vertex is at the same x-coordinate as the axis of symmetry, we can simply plug this value into our function to find the y-coordinate of the vertex:

$y=\left(\frac{3}{2}\right)^2-3\left(\frac{3}{2}\right)-5=\frac{9}{4}-\frac{9}{2}-5=-\frac{29}{4}$

So the vertex occurs at the point:

$\left(\frac{3}{2},-\frac{29}{4}\right)$

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Question

Find the axis of symmetry and the vertex of the parabola given by the following equation:

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Answer

Find the axis of symmetry and the vertex of the parabola given by the following equation:

To find the axis of symmetry of a parabola in standard form, , use the following equation:

$AoS=\frac{-b}{2a}$

So...

$AoS=\frac{-(-28)}{2\cdot2}=\frac{28}{4}=7$

This means that we have an axis of symmetry at . Or, to put it more plainly, at we could draw a vertical line which would perfectly cut our parabola in half!

So, we are halfway there, now we need the coordinates of our vertex. We already know the x-coordinate, which is 7. To find the y-coordinate, simply plug 7 into the parabola's formula and solve!

This makes our vertex the point

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Question

Write the standard equation of the circle.

$x^{2}+y^{2}+8x-4y-5=0$

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Answer

Group the x and y terms on one side of the equation and the constant on the other.

$(x^{2}+8x)+(y^{2}-4y)=5$

Complete the square by taking half of the middle number for each variable and squaring it. Add the number to the other side of the equation.

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

$(x^{2}+8x+16)+(y^{2}-4y+4)=5+16+4$

$(x+4)^{2}+(y-2)^{2}=5^{2}$

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Question

Find the directerix for the parabola with the following equation:

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Answer

Recall the standard form of the equation of a vertical parabola:

, where is the vertex of the parabola and gives the focal length.

When , the parabola will open up.

When , the parabola will open down.

For the parabola in question, the vertex is and . This parabola will open up. Because the parabola will open up, the directerix will be located unit down from the vertex. The equation for the directerix is then .

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Question

Find the foci of the hyperbola with the following equation:

$\frac{(x-3)^2}{45}-\frac{(y-2)^2}{4}=1$

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Answer

Recall that the standard formula of a hyperbola can come in two forms:

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

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}-=1$ , where the centers of both hyperbolas are .

When the term with is first, that means the foci will lie on a horizontal transverse axis.

When the term with is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

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

For a hyperbola with a horizontal transverse access, the foci will be located at and .

For a hyperbola with a vertical transverse access, the foci will be located at and .

For the given hypebola in the question, the transverse axis is horizontal and its center is located at .

Next, find .

$c=\sqrt{45+4}=\sqrt{49}=7$

The foci are then located at and .

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Question

Find the coordinate of intersection, if possible: and .

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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: ,

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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:

;

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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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Question

Give the coordinate pairs that satisfy the system of equations.

$\begin{matrix} y = 5x^2 - 10x + 6 \ y = 2x^2 + 9 x + 20 \end{matrix}$

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Answer

To solve, set the two quadratics equal to each other and then combine like terms:

subtract everything on the right from both sides to combine like terms.

Solving by factoring or using the quadratic formula gives us the solutions $-\frac{2}{3}$ and .

To find the y-coordinates, plug these into either equation:

$5\left( -\frac{ 2}{3} \right)^2 -10 \left(-\frac{ 2}{3}\right) + 6 = 14.\overline{8}$

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Question

Give the , coordinate pairs that satisfy the two equations.

$\begin{matrix} y = x^2 - 7x - 10 \ y +2x^2 + 18x = 10 \end{matrix}$

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Answer

To solve, first re-write the second one so that y is isolated on the left side:

Now set the two quadratics equal to each other:

add/subtract all of the terms from the right side so that this is a quadratic equal to zero.

combine like terms.

Using the quadratic formula or by factoring, we get the two solutions and $\frac{ 4}{3}$ .

To get the y-coordinates, plug these numbers into either function:

$y = \left(\frac{4}{3}\right)^2 - 7\left(\frac{ 4}{3} \right) - 10 = \frac {16}{9} - \frac{84}{9} - \frac{90}{9} = \frac{-158}{9 } = -17.\overline{5}$

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Question

Find the coordinate pairs satisfying both polynomials:

$\begin{matrix} y = 4x^3 + 3x^2 -10x - 5 \ y = 4x^3 + 2x^2 -8x + 10 \end{matrix}$

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Answer

To solve, set the two polynomials equal to each other:

add/subtract all of the terms from the right side from both sides.

combine like terms.

Solving with the quadratic formula or by factoring gives us the solutions 5 and -3.

To get the y-coordinates, plug these numbers into either of the original equations:

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Question

Which is the correct polar form of the cartesian equation $\frac{(y+1)^2 }{ 4 } + \frac{x^2 }{3 } = 1$ ?

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Answer

To determine the polar equation, first we need to interpret the original cartesian graph. This is an ellipse with a vertical major axis with half its length $a = \sqrt{4} = 2$ . The minor axis has half its length $b = \sqrt{3}$ . To find the foci, use the relationship

$\sqrt{3} ^ 2 = 2^ 2 - c ^ 2$

so

Since the center is , this means that the foci are at $(0, -1 \pm 1 ) = (0, -2); (0, 0 )$

Having a focus at the origin means we can use the formula $r = \frac{pe}{ 1 \pm e \cos \theta}$ \[or sine\] where e is the eccentricity, and for an ellipse $p = \frac{a^2 - c^2 }{ c }$ .

In this case, the focus at the origin is above the directrix, so we be subtracting. The major axis is vertical, so we are using sine.

Solving for p gives us: $p = \frac{2^2 - 1^2 }{ 1 } = 4 - 1 = 3$

The eccentricity is $e = \frac {c }{a }$ , in this case $\frac{1}{2 }$

This gives us an equation of:

$r = \frac{3\frac{1}{2}}{ 1 - \frac{1}{2} \sin \theta}$

We can simplify by multiplying top and bottom by 2:

$r = \frac{3\frac{1}{2}}{ 1 - \frac{1}{2} \sin \theta} * \frac{2}{2 } = \frac{ 3 }{ 2 - \sin \theta}$

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Question

Write the equation for the circle in polar form.

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Answer

To convert this cartesian equation to polar form, we will use the substitutions and $x = r \cos \theta$ .

First, we should expand the expression:

square x - 3

subtract 9 from both sides

group the squared variables next to each other - this will help see how to re-write this in polar form:

now make the substitutions:

$r^2 - 6 r \cos \theta = 0$

This is a quadratic in r with $a = 1, b = -6 \cos \theta, c = 0$

Solve using the quadratic formula:

$r = \frac {6 \cos \theta \pm \sqrt {(6 \cos \theta)^2 - 410}}{2*1} = \frac{6 \cos \theta \pm \sqrt{ (6 \cos \theta )^2 }}{2} = \frac {6 \cos \theta \pm 6 \cos \theta }{2}$

Subtracting $6 \cos \theta - 6 \cos \theta = 0$ which would give us a circle with no radius, but adding $6 \cos \theta + 6 \cos \theta = 12 \cos \theta$ so our answer is:

$r = \frac{12 \cos \theta }{2 } = 6 \cos \theta$

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