analytic geometry?

Question

What are the rectangular coordinates of the point (10,150*)

A.(5,-8.66)
B.(-5,-8.66)
C.(8.66,-5)
D.(-8.66,5)


2.Find the intersection points of the parabola
y²=-8x

and the circle x²+y²-2x-24=0

A.(2,-4)
B.(2,4)
C.(-2,4) and (-2,-4)
D.(2,4)and (-2,4)

Answer 1

1) i am guessing that these (10, 150°) are the POLAR coordinates of a point.,
then x= 10 cos(150°)
and
y=10sin(150°)

2) substitute -8x for y² in the equation of the circle:
x^2 -8x -2x -24 =0
x^2 -10x -24 =0
x= (10 +- sqrt( 100 -4(-24) ) /2
x= (10 +- 14)/2
x= (10+14)/2 =24/2 = 12, y²=-8(12) which does not have any solutions
or x = (10-14)/2 =-4/2 =-2, y² = -8(-2) = 16, so y = 4 or y =-4
so the solutions are:
C.(-2,4) and (-2,-4)
'

Answer 2

1) I take it the given coordinates are polar, with
r = 10 and theta = 150 deg.

The formula for rectangular coordinates is
(r* cos(theta), r* sin(theta)), so work it from there.

2) On the parabola, y²=-8x, so substitute -8x for y² in the equation of the circle, and then solve the quadratic equation you get to find x. Then sub the negative x value into the parabola equation to find two y values, and you have both points of intersection. The positive value of x (12) isn't possible because
y²=-8x would give a negative value for y².

Answer 3

1) You mean (10, 150°), a circular coordinate point? The angle is lways counted from the positive side of the x-axis (for example, (10, 0°) is x=10. y=0) and going counter-clockwise.

So 0° to 90° is in the first quadrant (x and y positive), etc. You can take it from there.

2) Replace y^2 with -8x in the second equation and there's your quadratic equation. Solve for x.

If you want to be real lazy, plug the multipple-choice answers into the equations to see if they work. But that's cheating...

Answer 4

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Answer 5

Three answers were all correct.These three are Hy, benoit353,and Polarbear.
You can plug in one by one to figure out which one is the answer.