Ask a New Question
Menu

Write the equation of the line tangent to the curve

x^2 − 4y^2 + 192 = 0 at (8, 8)
Thank You

1 answer

x^2 - 4y^2 + 1292 = 0
2x - 8yy' = 0
y' = x/4y

You have x and y, so y' = 1/4

the line is thus

y-8 = 1/4 (x-8)

see at

http://www.wolframalpha.com/input/?i=plot+x%5E2+%E2%88%92+4y%5E2+%2B+192+%3D+0%2C+y+%3D+1%2F4+%28x-8%29%2B8
Similar Questions
  1. original curve: 2y^3+6(x^2)y-12x^2+6y=1dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the
    1. answers icon 1 answer
  2. Consider the curve defined by 2y^3+6X^2(y)- 12x^2 +6y=1 .a. Show that dy/dx= (4x-2xy)/(x^2+y^2+1) b. Write an equation of each
    1. answers icon 3 answers
  3. original curve: 2y^3+6(x^2)y-12x^2+6y=1dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the
    1. answers icon 0 answers
  4. original curve: 2y^3+6(x^2)y-12x^2+6y=1dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the
    1. answers icon 3 answers
more similar questions