Question

Find the minimum distance from the origin to the parabola y=5-x^2

Answer #1

Find the equation of the osculating
circle of the parabola y =
x^2 at the
origin.

What is the distance between the point (1, 2) and the parabola y
= x^2.

Find the point with x ≥ 0 on the parabola y = (1/8)x^2 − 5 that
is closest to the point ( 0 , 1 ).

Find the perpendicular distance of the plane 5 x + 2 y – z = –22
from origin O by first
finding the co-ordinates of the point P on the plane such that OP
is perpendicular to the given plane.

a. Draw the parabola y=x^2 and the point (0,3) in the square
window -2 < x < 2 and 0 < y <4.
b. Fill in the four blanks to complete the formula
giving the distance D from the point (0,3) to a general point (x,y)
in the plane.
D = Sqrt[( - )^2 + ( - )^2]
c. Find the points on the parabola y=x^2 which are closest to the
point (0,3). You must have both appropriate calculations...

Find the minimum distance from the point (1,-6,3) to the plane x
− y + z = 7. (Hint: To simplify the computations, minimize
the square of the distance.)

graph the parabola y=-x^2 -4x +5. list vertex, aos, and y,x
intercepts

A particle with charge Q is on the y axis a distance A from the
origin and another particle with charge q is on the x axis a
distance D from the origin. The value of D for which the x
component of the force on the second article is the greatest
is:
A)0
B) A
C) Sqrt(2)A
D) A/2
E) A/sqrt 2

Find the x coordinate of the point, correct to two decimal
places, on the parabola y=6.17-x^2 at which the tangent line cuts
from the first quadrant the triangle with the smallest area.

There is a line through the origin that divides the region
bounded by the parabola y=8x−7x2 and the x-axis into two
regions with equal area. What is the slope of that line?

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