What is the equation to the tangent line to the curve x2 y2169 at the point 5-12?

Answer 1

first solve for y y= square root of (169-x2) find the first derivative y'=1/2 (1/sqr(169-x2))(2x) this gives you the formula for the slope. Use x=5 this gives the slope of 5/12. The formula for a straight line is y - y1 = m(x- x1), substitute the slope for m y + 12=(5/12)(x-5) use 5-12 for x and y 12y + 144 = 5x -25 5x- 12y = 169