Thread: Find the equation,

I thought this was a lil to simple and think my answer might be wrong :/

Determine the equation of the tangent to the curve at the given point

(x+2)^2+(y-3)^2=29 P(3,5)

Originally Posted by tnkfub

I thought this was a lil to simple and think my answer might be wrong :/

Determine the equation of the tangent to the curve at the given point

(x+2)^2+(y-3)^2=29 P(3,5)

The slope of the tangent line is dy/dx.

(x+2)^2 +(y-3)^2 = 29
Differentiate both sides with respect to x,
2(x+2) +2(y-3)(dy/dx) = 0
dy/dx = -(x+2)/(y-3) -----------***

At point (3,5),
dy/dx = -(3+2)/(5-3) = -5/2
Using the point-slope form of the equation of the tangent line,
y-5 = -(5/2)(x-3)

y = -(5/2)(x-3) +5
y = -(5/2)x +15/2 +5
y = -(5/2)x +25/2 ----------answer.
Or,
2y = -5x +25
5x +2y -25 = 0 -------answer.