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SAT II Math II : Coordinate Geometry

研究概念,例子问题&解释for SAT II Math II

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Example Questions

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Example Question #1 :Coordinate Geometry

Which of the following equations represent a parabola?

Possible Answers:

y^2 = x^2

$y=\frac{1}{2x}$

y=2^x

$y=\frac{1}{2x^2}$

$y=\frac{1}{2}x^2$

Correct answer:

$y=\frac{1}{2}x^2$

Explanation:

The parabola is represented in the form y=ax^2+bx+c ．If there is a variable in the denominator or as an exponent, it is not a parabola.

The only equation that has an order of two is: $y=\frac{1}{2}x^2$

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Example Question #1 :Circles, Ellipses, And Hyperbolas

Circle

Refer to the above figure. The circle has its center at the origin. What is the equation of the circle?

Possible Answers:

$x^{2}+ y^{2} = 100$

$x^{2}+ y^{2} = 1 0$

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

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

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

Correct answer:

$x^{2}+ y^{2} = 100$

Explanation:

The equation of a circle with center (h, k) and radiusis

$(x-h)^{2}+ \left ( y - k\right )^{2} = r^{2}$

The center is at the origin, or (0,0) , so h = k = 0 ．To find $r^{2}$ , use the distance formula as follows:

$r^{2} = \left ( 6-0\right )^{2} +\left ( 8 -0 \right )^{2} = 6^{2}+8^{2} = 36 + 64 = 100$

Note that we do not actually need to find．

We can now write the equation of the circle:

$(x-0)^{2}+ \left ( y - 0\right )^{2} = 100$

$x^{2}+y^{2} = 100$

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Example Question #2 :Circles, Ellipses, And Hyperbolas

Circle

Refer to the above diagram. The circle has its center at the origin;is the point (-9, -12) ．What is the length of the arc $\widehat{AB}$ , to the nearest tenth?

Possible Answers:

37.5

33.2

18.8

16.6

56.8

Correct answer:

33.2

Explanation:

First, it is necessary to determine the radius of the circle. This is the distance between (0,0) and (-9, -12) , so we apply the distance formula:

$r = \sqrt{(x_{2} - x _{1})^{2}+(y_{2} - y_{1})^{2}}$

$r = \sqrt{\left ( -9-0\right )^{2} +\left ( -12 -0 \right )^{2} }$

$r=\sqrt{ (-9)^{2}+(-12)^{2}}$

$r= \sqrt{81+144 }= \sqrt{225} = 15$

The circumference of the circle is

$C = 2 \pi r = 2 \pi \cdot 15 = 30 \pi$

Now we need to find the degree measure of the arc. We can do this best by examining this diagram:

Circle

The degree measure of $\widehat{AB}$ is also the measure $\theta$ of the central angle formed by the green radii. This is found using the relationship

$\tan \theta = \frac{y}{x}$

$\tan \theta = \frac{-12}{-9} \approx 1.333$

Using a calculator, we find that $\tan ^{-1} 1.3333 \approx 53.12^{\circ }$ ．We can adjust for the location of:

$\theta \approx \left (180 - 53.12 \right )^{\circ } \approx 126.87^{\circ }$

which is the degree measure of the arc.

Now we can calculate the length of the arc:

$L \approx \frac{126.87}{360} \cdot C$

$L \approx \frac{126.87}{360} \cdot 30 \pi \approx 33.2$

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Example Question #3 :Circles, Ellipses, And Hyperbolas

On the coordinate plane, the vertices of a square are at the points with coordinates (7, 0), (0, 7), (-7, 0) , (0, -7) ．Give the equation of the circle.

Possible Answers:

$x^{2}+ y^{2}= \frac{7}{2}$

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

$x^{2}+ y^{2}= \frac{49}{2}$

$(x- 7) ^{2}+ (y- 7) ^{2}= \frac{49}{4}$

$x^{2}+ y^{2}= \frac{49}{4}$

Correct answer:

$x^{2}+ y^{2}= \frac{49}{2}$

Explanation:

The figure in question is below.

Incircle 1

The center of the circle can be seen to be the origin, so, if the radius is, the equation will be

$x^{2}+ y^{2} = r^{2}$ ．

The circle passes through the midpoints of the sides, so we will find one of these midpoints. The midpoint $(x_{m}, y_{m})$ of the segment with endpoints (7,0) and (0, 7) can be found by using the midpoint equations, setting $x_{1} = y_{2}= 7, x_{2} = y_{1}= 0$ :

$x_{m} = \frac{x_{1}+x_{2}}{2} = \frac{7+0}{2} = \frac{7}{2}$

$y_{m} = \frac{y_{1}+y_{2}}{2} = \frac{0+7 }{2} = \frac{7}{2}$

The circle passed though this midpoint $\left ( \frac{7}{2}, \frac{7}{2} \right )$ ．The segment from this point to the origin (0,0) is a radius, and its length is equal to．Using the following form of the distance formula, since we only need the square of the radius:

$r ^{2} = (x_{2}-x_{1}) ^{2} + (y_{2}-y_{1}) ^{2}$ ,

set $x_{1} = y_{1} = 0, x_{2}= y_{2}= \frac{7}{2}$ :

$r ^{2} =\left (\frac{7}{2}-0\right ) ^{2}+\left (\frac{7}{2}-0\right ) ^{2}$

$=\left (\frac{7}{2} \right ) ^{2}+\left (\frac{7}{2}\right ) ^{2}$

$= \frac{49}{4} + \frac{49}{4}$

$= \frac{49}{2}$

Substituting in the circle equation for $r ^{2}$ , we get the correct response,

$x^{2}+ y^{2}= \frac{49}{2}$

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Example Question #4 :Circles, Ellipses, And Hyperbolas

Find the diameter of the circle with the equation x^2-10x+y^2+2y+10=0 ．

Possible Answers:

Correct answer:

Explanation:

Start by putting the equation in the standard form of the equation of a circle by completing the square. Recall the standard form of the equation of a circle:

(x-a)^2+(y-b)^2=r^2 , where the center of the circle is at (a,b) and the radius is．

x^2-10x+25+y^2+2y+1+10=25+1

(x-5)^2+(y+1)^2+10=26

(x-5)^2+(y+1)^2=16

From the equation, we know that r^2=16 ．

$r=\sqrt{16}=4$

Since the radius is, double its length to find the length of the diameter. The length of the diameter is．

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Example Question #1 :Circles, Ellipses, And Hyperbolas

A triangle has its vertices at the points with coordinates (0,0) , (4, 4) , and (7, 0 ) ．给circumscrib的圆的方程es it.

Possible Answers:

$x^{2}+y^{2}-7x-y=0$

$x^{2}+y^{2}+7x-y=0$

$x^{2}+ y^{2}+7x -15y = 0$

None of these

$x^{2}+y^{2}-7x-15 y=0$

Correct answer:

$x^{2}+y^{2}-7x-y=0$

Explanation:

The circumscribed circle of a triangle is the circle which passes through all three vertices of the triangle.

In general form, the equation of a circle is

$x^{2}+ y^{2}+ Ax + By + C = 0$ ．

Since the circle passes through the origin, substitute x = 0, y = 0 ; the equation becomes

$0^{2}+0^{2}+ A \cdot 0 + B \cdot 0 + C = 0$

0+0+0+0 + C = 0

C= 0

Therefore, we know the equation of any circle passing through the origin takes the form

$x^{2}+ y^{2}+ Ax + By = 0$

for some A , B ．

Since the circle passes through (7,0) , substitute x = 7, y = 0 ; the equation becomes

$7^{2}+ 0^{2}+ A \cdot 7 + B \cdot 0 = 0$

49+7A= 0

Solving for:

7A = -49

A = -7

Now we know that the equation takes the form

$x^{2}+ y^{2}-7x + By = 0$

for some．

Since the circle passes through (4, 4) , substitute x = 7, y = 0 ; the equation becomes

$4^{2}+ 4^{2}- 7 \cdot 4 + B \cdot 4 = 0$

16+16- 28 + 4B = 0

4 + 4B = 0

Solving for:

4B = -4

B = -1

The general form of the equation of the circle is therefore

$x^{2}+ y^{2}-7x -y = 0$

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Example Question #1 :Symmetry

Which of the following symmetries applies to the graph of the relation

$\left (x- 4 \right )^{2} + y ^{2}= 18$ ?

I) Symmetry with respect to the origin

II) Symmetry with respect to the-axis

III) Symmetry with respect to the-axis

Possible Answers:

II only

I, II, and III

III only

I only

None of these

Correct answer:

II only

Explanation:

The relation

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

is a circle with center (h,k) and radius．

In other words, it is a circle with center at the origin, translated rightunits and upunits.

$\left (x- 4 \right )^{2} + y ^{2}= 18$

$\left (x- 4 \right )^{2} +\left ( y - 0 \right ) ^{2}= 18$

is a circle translated right 4 units and up zero units. The upshot is that the circle moves along the-axis only, and therefore is symmetric with respect to the-axis, but not the-axis. Also, as a consequence, it is not symmetric with respect to the origin.

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Example Question #1 :Transformations

Axes_1

Refer to the above figure.

Which of the following functions is graphed?

Possible Answers:

g(x) = |x - 6| + 3

g(x) = |x+3| + 6

The correct answer is not given among the other responses.

g(x) = |x-3| + 6

g(x) = |x +6| + 3

Correct answer:

g(x) = |x - 6| + 3

Explanation:

Below is the graph of f(x) = |x| :

Axes_1

The given graph is the graph ofshifted 6 units left (that is,unit right) and 3 units up.

The function graphed is therefore

g (x) = f(x-h) + k where h = 6, k= 3 ．That is,

g (x) = f(x-6) +3

g(x)= | x-6 | + 3

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Example Question #2 :Transformations

Axes_1

Refer to the above figure.

Which of the following functions is graphed?

Possible Answers:

$g(x) = \left |3x + \frac{3}{2} \right |\textup{}$

g(x) = |3x +6|

g(x) = |2x| + 3

g(x) = |2x + 6|

$g(x) =\left | \frac{1}{2} x \right |+ 3$

Correct answer:

g(x) = |2x| + 3

Explanation:

Below is the graph of f(x) = |x| :

Axes_1

If the graph ofis translated by shifting each point (x,y) to the point (2, 2y) , the graph of

g(x) = 2f(x)

就形成了。如果图向上移three units, the new graph is

g(x) = 2f(x) + 3

Since the starting graph was f(x) = |x| , the final graph is

g(x) = 2 |x| + 3 , or,

g(x) = |2x| + 3

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Example Question #3 :Transformations

Axes_1

Refer to the above figure.

Which of the following functions is graphed?

Possible Answers:

g(x) = |3x-12|

$g(x) =\left | \frac{1}{3} x-\frac{4}{3} \right |$

g(x) =- |3x+12|

$g(x) =\left | \frac{1}{3} x+\frac{4}{3} \right |$

g(x) = |3x+12|

Correct answer:

g(x) = |3x-12|

Explanation:

Below is the graph of f(x) = |x| :

Axes_1

If the graph ofis translated by shifting each point (x,y) to the point (x, 3y) , the graph of

g(x) = 3f(x)

就形成了。If the graph is then shifted right by four units, the new graph is

g(x) = 3f(x- 4)

Since the starting graph was f(x) = |x| , the final graph is

g(x) = 3 |x-4| , or g(x) = |3x-12|

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SAT II Math II : Coordinate Geometry

研究概念,例子问题&解释for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #1 :Coordinate Geometry

Example Question #1 :Circles, Ellipses, And Hyperbolas

Example Question #2 :Circles, Ellipses, And Hyperbolas

Example Question #3 :Circles, Ellipses, And Hyperbolas

Example Question #4 :Circles, Ellipses, And Hyperbolas

Example Question #1 :Circles, Ellipses, And Hyperbolas

Example Question #1 :Symmetry

Example Question #1 :Transformations

Example Question #2 :Transformations

Example Question #3 :Transformations

All SAT II Math II Resources

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