What is the equation of the line tangent to the curve x^2+y^2=9 at (0,3)?

Answer 1

Basically you have a circle centered at the origin of radius 3. One can visually realize that the slope at 0,3 is 0, or a horizontal line. But to solve it, you need to use implicit differentiation.

x² + y² = 9 ⇒ x² + y² - 9 = 0

D_x(x² + y² - 9) ⇒ D_x(x²) + D_x(y²) - D_x(9) ⇔ 2x + 2yy' + 0

Solving for y' we get 2yy' = -2x ⇒ y' = -2x/2y = -x/y

At the point (0,3) we see that the slope is -0/y = 0.

Notice that if the point was (3,0) the slope would be undefined, which is what you get from a vertical slope. Draw a circle, radius 3, and put a line at 0,3 and at 3,0 to see.

Answer 2

This is an ellipse and you are asked for the tangent at one of the y-intercepts.

No calculus is needed. The answer is y=3.