Posted by **L.Bianchessi** on Friday, March 9, 2012 at 7:41pm.

Let P(x,y) be a point on the curve y=16-x^2 and let A be the point directly below P on the x axis. If O is the origin, find the area and the perimeter of triangle OAP as functions of x.

I got Area=x(8-1/2x^2). Is that right?

And perimeter= x+16-x^2+ (square root 256-30x^2)

Did I do this question right?

- MaTH!!! -
**Reiny**, Friday, March 9, 2012 at 9:17pm
Your area is correct

But for the perimeter, ....

let the hypotenuse by y

y^2 = x^2 + (16-x^2)^2

= x^2 + 256 - 32x^2 + x^4

y = √(256 - 31x^2 + x^4)

perimeter = x + √(256-31x^2+x^4) + 16- x^2

testing:

let P be (3, 7)

then AP = 7

OA = 3 and OP = √(9+49) = √58

Perimeter = 3 + 7 + √58 = 10+ √58

according to my formula

Perimeter = 3 + √(256 - 31(9) + 81) + 16 - 9

= 10 + √58

Well, how about that?

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