Question
Find the tangent line to the curve (x^2/16)-(y^2/9)=1 at the point (4sqrt2,3)
Answers
Reiny
x/8 - (2/9)y dy/dx = 0
dy/dx = (x/8) / (2y/9) = 9x/(16y)
at (4?2,3), dy/dx = 36?2/(48) = 3?2/4
equation:
y - 3 = (3?2/4)(x - 4?2)
clean it up to whatever form is needed.
Proof:
http://www.wolframalpha.com/input/?i=plot+(x%5E2%2F16)-(y%5E2%2F9)%3D1+,+y+-+3+%3D+(3%E2%88%9A2%2F4)(x+-+4%E2%88%9A2)
dy/dx = (x/8) / (2y/9) = 9x/(16y)
at (4?2,3), dy/dx = 36?2/(48) = 3?2/4
equation:
y - 3 = (3?2/4)(x - 4?2)
clean it up to whatever form is needed.
Proof:
http://www.wolframalpha.com/input/?i=plot+(x%5E2%2F16)-(y%5E2%2F9)%3D1+,+y+-+3+%3D+(3%E2%88%9A2%2F4)(x+-+4%E2%88%9A2)