Calculus: Parabola and Straight Line?

Question

Need someones help on solving this problem: Show parabola y = -x^2 and line x-4y-18=0 intersect at right angles at one of their points of intersection.

Answer 1

1.) Find the intersection points.
y= x^2
x - 4y - 18 = 0
To do so, set both equations to equal y, at which point they equal each other.
For equation #2:
x - 4y - 18 = 0
x - 18 = 4y
y = 1/4x - 18/4

Set the equations to equal each other:
-x^2 = 1/4x - 18/4
x^2 + 1/4x - 18/4 = 0

Now you have a quadratic equation...so use the quadratic formula:
x = (-b +/- (b^2 - 4ac)^1/2) / 2a
x = (-1/4 +/- (1/16 - (4 * - 18/4 * 1))^1/2) / 2
x = (-1/4 +/- (1/16 + 18)^1/2) / 2
x = -1/8 +/- ((289 / 16)^1/2) / 2
x = -1/8 +/- (17/4) / 2 = -1/8 +/- 17/8
x = -18/8, 16/8
x = -9/4, 2

To find the slope of a parabola at any point, take the derivative:
y = -x^2
y' (and thus, slope) = -2x

At -9/4, the slope is 9/2
At 2, the slope is -4

Now, compare the slope of the parabola at the 2 intersection points with the slope of your line:
y = 1/4x - 18/4, therefore the slope = 1/4

2 lines are perpendicular if the slopes are negative reciprocals of each other, therefore, the line is perpendicular to the parabola at the intersection at (2, -4).

Answer 2

You'll need to determine the slope of the parabola at the point of intersection. (You should be able to see the slope of the line quite easily.) If the parabola slope is the negative inverse of the line slope, you're golden.

To get the parabola slope, you'll need to take the derivative at the point of intersection (because the parabola is never truly straight, even far from its vertex where it looks straight). Use your text to figure out how.

Answer 3

in theory, you should find the intersection point (x,y) then you take the first dirivative of y=-x^2 (y'=-2x) and prove that it is equal to -4 (perpendicular slope to the +1/4 slope of the straight line) at the x value where the two graphs intersect. unfortunately, it doesnt look like these two graphs actually intersect at all... at least in the real number plane. sorry I can't be of more help.

Answer 4

dy/dx = 4 - 2x Let the point of tangency be (a, b). m = 4 - 2a (5 - b)/(2 - a) = 4 - 2a b = 4a - a^2 5 - b = (4 - 2a)(2 - a) = 2(2 - a)^2 = 2(4 - 4a + a^2) = 8 - 8a + 2a^2 -3 + 8a - 2a^2 = b -3 + 8a - 2a^2 = 4a - a^2 0 = a^2 - 4a + 3 0 = (a - 1)(a - 3) a = 1 or a = 3 b = 3 or b = 3 (1, 3) or (3, 3) y = 2x + 1 and y = -2x + 9

Answer 5

Its the point (2, -4)
The other point is diefinatly not a right angle, it doesn't even look like one. Hope that is the right answer.

Answer 6

wit those two lines they only intercept at 2 places, therefore there is no question and the interception probly isnt at right angles....i dont understand!

Answer 7

Sorry bro I have no clue.