How to find the equation of a curve - SSAT Upper Level Quantitative

Card 0 of 76

Question

An ellipse passes through points .

Give its equation.

Answer

The equation of the ellipse with center , horizontal axis of length , and vertical axis of length is

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

and are the endpoints of a horizontal line segment with midpoint

$\left ( \frac{4+10}{2}, \frac{6+6}{2} \right )$ , or

and length .

and are the endpoints of a vertical line segment with midpoint

$\left ( \frac{7+7}{2}, \frac{2+10}{2} \right )$ , or

and length

Because their midpoints coincide, these are the endpoints of the horizontal axis and vertical axis, respectively, of the ellipse, and the common midpoint is the center.

Therefore,

and ;

and ; consequently and .

The equation of the ellipse is

$\frac{(x-7)^{2}}{3^{2}} + \frac{(y-6)^{2}}{4^{2}} = 1$ , or

$\frac{(x-7)^{2}}{9} + \frac{(y-6)^{2}}{16} = 1$

Compare your answer with the correct one above

Question

If the -intercept of the line is and the slope is , which of the following equations best satisfies this condition?

Answer

Write the slope-intercept form.

The point given the x-intercept of 6 is .

Substitute the point and the slope into the equation and solve for the y-intercept.

Substitute the y-intercept back to the slope-intercept form to get your equation.

Compare your answer with the correct one above

Question

A vertical parabola on the coordinate plane includes points and .

Give its equation.

Answer

The standard form of the equation of a vertical parabola is

$y=ax^{2}+bx+c$

If the values of and from each ordered pair are substituted in succession, three equations in three variables are formed:

$a\cdot (-2)^{2}+b(-2)+c = 23$

$a\cdot 1^{2}+b \cdot 1+c = 5$

$a\cdot 3^{2}+b \cdot 3+c =13$

The system

can be solved through the elimination method.

First, multiply the second equation by and add to the third:

$\underline{ -a-b-c ; ; ; =- 5 }$

Next, multiply the second equation by and add to the first:

$\underline{ -a-b-c ; =- 5 }$

Which can be divided by 3 on both sides to yield

Now solve the two-by-two system

by substitution:

Back-solve:

Back-solve again:

The equation of the parabola is therefore

$y = 2x^{2}-4x+7$ .

Compare your answer with the correct one above

Question

A vertical parabola on the coordinate plane has vertex and -intercept .

Give its equation.

Answer

The equation of a vertical parabola, in vertex form, is

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

where is the vertex. Set :

$y = a[x-(-4)]^{2}+10$

$y = a(x+4)^{2}+10$

To find , use the -intercept, setting :

$-2 = a(0+4)^{2}+10$

$-2 = a\cdot 4^{2}+10$

$a =- \frac{3}{4}$

The equation, in vertex form, is $y =- \frac{3}{4}(x+4)^{2}+10$ ; in standard form:

$y = -\frac{3}{4}(x+4)^{2}+10$

$y =- \frac{3}{4}(x^{2} + 8x+16)+10$

$y = -\frac{3}{4} x^{2}- 6x-12+10$

$y = -\frac{3}{4} x^{2}- 6x-2$

Compare your answer with the correct one above

Question

A vertical parabola on the coordinate plane has vertex ; one of its -intercepts is .

Give its equation.

Answer

The equation of a vertical parabola, in vertex form, is

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

where is the vertex. Set :

$y = a[x-(-3)]^{2}+ (-6)$

$y = a(x+3)^{2}-6$

To find , use the known -intercept, setting :

$0 = a(-7+3)^{2}-6$

$0 = a(-4)^{2}-6$

$a = \frac{6}{16} = \frac{3}{8}$

The equation, in vertex form, is $y = \frac{3}{8}(x+3)^{2}-6$ ; in standard form:

$y = \frac{3}{8}(x+3)^{2}-6$

$y = \frac{3}{8}(x^{2}+6x+9)-6$

$y = \frac{3}{8}x^{2}+ \frac{9}{4}x+ \frac{27}{8}-6$

$y = \frac{3}{8}x^{2}+ \frac{9}{4}x+ \frac{27}{8}- \frac{48}{8}$

$y = \frac{3}{8}x^{2}+ \frac{9}{4}x - \frac{21}{8}$

Compare your answer with the correct one above

Question

A vertical parabola on the coordinate plane has -intercept ; its only -intercept is .

Give its equation.

Answer

If a vertical parabola has only one -intercept, which here is , that point doubles as its vertex as well.

The equation of a vertical parabola, in vertex form, is

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

where is the vertex. Set :

$y = a(x-6)^{2}+0$

$y = a(x-6)^{2}$

To find , use the -intercept, setting :

$4= a(0-6)^{2}$

$4= a( -6)^{2}$

$a = \frac{4}{36} = \frac{1}{9}$

The equation, in vertex form, is $y = \frac{1}{9}(x-6)^{2}$ . In standard form:

$y = \frac{1}{9}(x-6)^{2}$

$y = \frac{1}{9}(x^{2}-12+36)$

$y = \frac{1}{9} x^{2}-\frac{4}{3}x+4$

Compare your answer with the correct one above

Question

A vertical parabola on the coordinate plane has -intercept ; one of its -intercepts is .

Give its equation.

Answer

The equation of a vertical parabola, in standard form, is

$y = ax^{2}+ bx + c$

for some real .

is the -coordinate of the -intercept, so , and the equation is

$y = ax^{2}+ bx + 6$

Set :

$0 = a\cdot 2^{2}+ b \cdot 2 + 6$

However, no other information is given, so the values of and cannot be determined for certain. The correct response is that insufficient information is given.

Compare your answer with the correct one above

Question

A vertical parabola on the coordinate plane has -intercepts and , and passes through .

Give its equation.

Answer

A vertical parabola which passes through and has as its equation

To find , substitute the coordinates of the third point, setting :

$-2 = a \cdot 1 \cdot (-3)$

$a = \frac{-2}{-3}= \frac{2}{3}$

The equation is $y = \frac{2}{3}(x-2)(x-6)$ ; expand to put it in standard form:

$y = \frac{2}{3}(x^{2}-8x+12)$

$y = \frac{2}{3}x^{2}-\frac{16}{3}x+8$

Compare your answer with the correct one above

Question

An ellipse on the coordinate plane has as its center the point . It passes through the points and . Give its equation.

Answer

The equation of the ellipse with center , horizontal axis of length , and vertical axis of length is

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

The center is , so and .

To find , note that one endpoint of the horizontal axis is given by the point with the same -coordinate through which it passes, namely, . Half the length of this axis, which is , is the difference of the -coordinates, so . Similarly, to find , note that one endpoint of the vertical axis is given by the point with the same -coordinate through which it passes, namely, . Half the length of this axis, which is , is the difference of the -coordinates, so .

The equation is

$\frac{(x-(-4))^{2}}{15^{2}} + \frac{(y-9)^{2}}{4^{2}} = 1$

or

$\frac{(x+4)^{2}}{225} + \frac{(y-9)^{2}}{16} = 1$ .

Compare your answer with the correct one above

Question

A vertical parabola on the coordinate plane shares one -intercept with the line of the equation , and the other with the line of the equation . It also passes through . Give the equation of the parabola.

Answer

First, find the -intercepts—the points of intersection with the -axis—of the lines by substituting 0 for in both equations.

is the -intercept of this line.

is the -intercept of this line.

The parabola has -intercepts at and , so its equation can be expressed as

for some real . To find it, substitute using the coordinates of the third point, setting :

$-4 = a \cdot 3 \cdot 2$

$a = \frac{-4}{6} = -\frac{2}{3}$ .

The equation is $y = -\frac{2}{3}(x-3)(x-4)$ , which, in standard form, can be rewritten as:

$y = -\frac{2}{3}(x^{2}-7x+12)$

$y = -\frac{2}{3}x^{2}+\frac{14}{3}x-8$

Compare your answer with the correct one above

Question

The -intercept and the only -intercept of a vertical parabola on the coordinate plane coincide with the -intercept and the -intercept of the line of the equation . Give the equation of the parabola.

Answer

To find the -intercept, that is, the point of intersection with the -axis, of the line of equation , set and solve for :

The -intercept is .

The -intercept can be found by doing the opposite:

The -intercept is .

The parabola has these intercepts as well. Also, since the vertical parabola has only one -intercept, that point doubles as its vertex as well.

The equation of a vertical parabola, in vertex form, is

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

where is the vertex. Set :

$y = a(x-10)^{2}+0$

$y = a(x-10)^{2}$

for some real . To find it, use the -intercept, setting

$6 = a(0-10)^{2}$

$6 = a( -10)^{2}$

$a = \frac{6}{100} = \frac{3}{50}$

The parabola has equation $y =\frac{3}{50} (x-10)^{2}$ , which is rewritten as

$y =\frac{3}{50} (x^{2}-20x+100)$

$y = \frac{3}{50} x^{2}-\frac{6}{5}x+6$

Compare your answer with the correct one above

Question

Give the equation of the above ellipse.

Answer

The equation of the ellipse with center , horizontal axis of length , and vertical axis of length is

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

The ellipse has center , horizontal axis of length 8, and vertical axis of length 6. Therefore,

, $a = \frac{8}{2} = 4$ , and $b = \frac{6}{2} = 3$ .

The equation of the ellipse is

$\frac{(x-3)^{2}}{4^{2}} + \frac{(y-5)^{2}}{3^{2}} = 1$

$\frac{(x-3)^{2}}{16} + \frac{(y-5)^{2}}{9} = 1$

Compare your answer with the correct one above

Question

Give the equation of the above ellipse.

Answer

The equation of the ellipse with center , horizontal axis of length , and vertical axis of length is

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

The ellipse has center , horizontal axis of length 8, and vertical axis of length 16. Therefore,

, $a = \frac{8}{2} = 4$ , and $b = \frac{16}{2} = 8$ .

The equation of the ellipse is

$\frac{(x-6)^{2}}{4^{2}} + \frac{(y-2)^{2}}{8^{2}} = 1$

$\frac{(x-6)^{2}}{16} + \frac{(y-2)^{2}}{64} = 1$

Compare your answer with the correct one above

Question

Give the equation of the above ellipse.

Answer

The equation of the ellipse with center , horizontal axis of length , and vertical axis of length is

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

The ellipse has center , horizontal axis of length 10, and vertical axis of length 6. Therefore,

, $a = \frac{10}{2} = 5$ , and $b = \frac{6}{2} = 3$ .

The equation of the ellipse is

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

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

Compare your answer with the correct one above

Question

A horizontal parabola on the coordinate plane as its only -intercept; its -intercept is .

Give its equation.

Answer

If a horizontal parabola has only one -intercept, which here is , that point doubles as its vertex as well.

The equation of a horizontal parabola, in vertex form, is

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

where is the vertex. Set :

$x = a(y-4)^{2}+0$

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

To find , use the -intercept, setting :

$6 = a(0-4)^{2}$

$6 = a( -4)^{2}$

$a = \frac{6}{16} = \frac{3}{8}$

The equation, in vertex form, is $x = \frac{3}{8}(y-4)^{2}$ . In standard form:

$x = \frac{3}{8}(y^{2}-8y+16)$

$x = \frac{3}{8} y^{2}-3y+6$

Compare your answer with the correct one above

Question

A horizontal parabola on the coordinate plane has -intercept ; one of its -intercepts is .

Give its equation.

Answer

The equation of a horizontal parabola, in standard form, is

$x = ay^{2}+ by + c$

for some real .

is the -coordinate of the -intercept, so , and the equation is

$x = ay^{2}+ by + 6$

Set :

$0 = a\cdot 2^{2}+ b \cdot 2 + 6$

However, no other information is given, so the values of and cannot be determined for certain. The correct response is that insufficient information is given.

Compare your answer with the correct one above

Question

A horizontal parabola on the coordinate plane has vertex ; one of its -intercepts is .

Give its equation.

Answer

The equation of a horizontal parabola, in vertex form, is

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

where is the vertex. Set :

$x = a[y-(-6)]^{2}+ (-3)$

$x = a(y+6)^{2}-3$

To find , use the known -intercept, setting :

$0 = a(4+6)^{2}-3$

$0 = a\cdot 10^{2}-3$

$a = \frac{3}{100}$

The equation, in vertex form, is $x = \frac{3}{100}(y+6)^{2}-3$ ; in standard form:

$x = \frac{3}{100}(y^{2}+12+36)-3$

$x = \frac{3}{100} y^{2}+\frac{9}{25} y+\frac{27}{25} - \frac{75}{25}$

$x = \frac{3}{100} y^{2}+\frac{9}{25} y-\frac{48}{25}$

Compare your answer with the correct one above

Question

A horizontal parabola on the coordinate plane includes points , and .

Give its equation.

Answer

The standard form of the equation of a horizontal parabola is

$x=ay^{2}+by+ c$

If the values of and from each ordered pair are substituted in succession, three equations in three variables are formed:

$a \cdot (-1)^{2}+b \cdot (-1)+ c = -11$

$a \cdot (1)^{2}+b \cdot 1+ c = -1$

$a \cdot (2)^{2}+b \cdot 2+ c = 10$

The three-by-three linear system

can be solved by way of the elimination method.

can be found first, by multiplying the first equation by and add it to the second:

$\underline{-a+b - c = 11}$

$2b \div 2 = 10 \div 2$

Substitute 5 for in the last two equations to form a two-by-two linear system:

$4a+2 \cdot 5+ c = 10$

The system

can be solved by way of the substitution method;

$-3a\div (-3) = -6 \div (-3)$

Substitute 2 for in the top equation:

The equation is $x=2y^{2}+5y-8$ .

Compare your answer with the correct one above

Question

A vertical parabola on the coordinate plane has -intercepts and , and passes through .

Give its equation.

Answer

A horizontal parabola which passes through and has as its equation

.

To find , substitute the coordinates of the third point, setting :

$a = \frac{3}{32}$

The equation is therefore $x = \frac{3}{32}(y-2)(y-6)$ , which is, in standard form:

$x = \frac{3}{32}(y^{2}-8y+12)$

$x = \frac{3}{32} y^{2}- \frac{3}{4} y+ \frac{9}{8}$

Compare your answer with the correct one above

Question

If the -intercept of the line is and the slope is , which of the following equations best satisfies this condition?

Answer

Write the slope-intercept form.

The point given the x-intercept of 6 is .

Substitute the point and the slope into the equation and solve for the y-intercept.

Substitute the y-intercept back to the slope-intercept form to get your equation.

Compare your answer with the correct one above

Tap the card to reveal the answer