help with tangent/normal to implicit curves problem??

Question

it asks.... find equations for the lines that ar tangent and normal to the curve at the given point

1. x² + 2y² = 9 , (1,2)

help!

Answer 1

The slope of the tangent line is equal to the value of the curve's first derivative at the tangent point.

To find dy/dx, take the derivative of each term in the curve expression with respect to x:

2x + 4y(dy/dx) = 0

dy/dx = - 2x/4y = - x/2y

slope of tangent line = m = dy/dx @ point (1,2) = -1/4

The simplest way to represent a line is with the slope-intercept form, y = mx + b. We still need to find b, which is easy now that we have a slope and a point to work with.

b = y - mx = 2 - (-1/4)*(1) = 2 + 1/4 = 9/4

The equation of the tangent line is:

y = -x/4 + 9/4 = (9 - x)/4

Answer 2

y = (x^9 - 4) * (x + 6y) dy = (x^9 - 4) * (dx + 6dy) + (x + 6y) * 9x^8 * dx dy = 6 * (x^9 - 4) * dy - (x^9 - 4) * dx + (x + 6y) * 9x^8 * dx x = a million y = -3/19 dy = 6 * (a million - 4) * dy - (a million - 4) * dx + (a million - 18/19) * 9 * a million * dx dy = -18dy + 3dx + 9dx/19 dy + 18dy = (fifty seven + 9) * dx / 19 19 * dy = (sixty 8/19) * dx dy / dx = sixty 8 / 361