The point on the curve?

Question

The point on the curve x^2+2y=0 that is nearest the point (0, -1/2) occurs where y is 

 

a) 1/2

b) 0

c) -1/2

d) -1

e) none of the above

 

Answer: B

 

Please show all your work. Thanks. 

Answer 1

The point on the curve x^2+2y=0 that is nearest the point (0, -1/2) occurs where y is

a) 1/2
b) 0
c) -1/2
d) -1
e) none of the above

Answer: B

x^2+2y=0 ==> y = –1/2 x^2

Which is a parabola opening down with vertex at (0,0).

(0,-1/2) is 1/2 unit below the graph’s vertex, so we might suspect that the nearest point is the vertex. Let’s prove it.

Point on curve: (x,–1/2 x^2)

Distance between points = d.

d^2 = (x-0)^2 + (-1/2 + 1/2 x^2)^2

= x^2 + 1/4 (x^2-1)^2

4d^2 = 4x^2 + x^4 - 2x^2 + 1

= x^4 + 2x^2 + 1 = (x^2+1)^2

2d = x^2+1

The minimum distance occurs when x = 0 and d = 1/2.

So our conjecture was true.

Then answer is B, y = –1/2 0^2 = 0.

Answer 2

You need to use the values given and substitute them into the problem. When X=0 what is Y? When Y=-1/2 what is x. Which of those points are closest to (0,-1/2).

 

Conversely, You can evaluate each answer and find out which one is closest to the point, i.e. When Y=1/2 what is X?