2x - 8yy' = 0
y' = x/4y
Now we have a problem. The y 9s not defined for x=1. (y^2 would have to be negative.)
However, if you fix that, just find y(1) to get a point on the curve. Then the slope there is y'(1).
the line is thus
y-y(1) = y'(1) (x-1)
A function is implicitly defined
by x^2 - 4y^2 = 9
Find the equation of the tangent to the function at x = 1
1 answer
1 answer