Remember that = 1, which gives us
2 x − + 8y = 0Step 2: Now just simplify and solve for .
2 x − 2x − 2y + 8y = 0x − 2x − y + 4y = 0− x + 4y = y − x(4y − x) = y − x(b) Find the equation of the tangent lines to the curve at the point x = 2.
Step 1: We are going to use the point-slope form of a line, y − y 1 = m(x − x 1 ), where (x 1 , y 1 ) is
a point on the curve, and the derivative at that point is the slope m. First, we need to know the
value of y when x = 2. If we plug 2 for x into the original equation, we get4 − 4y + 4y^2 = 524 y^2 − 4y − 48 = 0Divide through by 4: y^2 − y − 12 = 0
Factor: (y − 4)(y + 3) = 0, so y = 4 or y = −3.
Notice that there are two values of y when x = 2, which is why there are two tangent lines.
Step 2: Now that we have our points, we need the slope of the tangent line at x = 2.