Question

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

Answer #1

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...

1. Find the point on the curve y = x2 that is closest to (0,
5).
2. Find the function f(x),iff′′(x)=sinx+x and f(0)=f(π)=0.
3. Find derivatives of the following functions. a) arcsin(
square root 3x)

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

consider the region r bounded by the parabola y=4x^2 and the
lines x=0 and y=16 find the volume of the solid obtained by
revolving R about the line x=1

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

Find the point(s) of intersection, if any, of the line
x-2/1 = y+1/-2 = z+3/-5 and the plane 3x + 19y - 7a - 8 =0

1. (a) If a chord of the parabola y
2 = 4ax is a normal at one of its ends, show that its
mid-point
lies on the curve
2(xx 2a) = y
2
a
+
8a
3
y
2
.
Prove that the shortest length of such a chord is 6a√3
(b) Find the asymptotes of the hyperbola
x
2
2y
2 + 2x + y + 9 = 0.

Consider the parabola given by the equation
y = x2 −
5
4
. Let
(a,b)
be the point lying on this parabola, with
a ≥ 0
, which is closest to the origin.
What is
a2 / b2
?
(The solution, as usual, is a whole number.)

The jpdf of X and Y is
?(1/8)xe−(x+y)/2 if x > 0 and y > 0,
0 elsewhere.
Find E(Y/X).

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