Ellipse Calculus Question

Question

at what point is the slope of the tangent line to the ellipse(x^2/25+y^2/4=1) equal to =1

b)what is the lowest the x coordinate (x,y), what is the largest the x coordinate (x,y)

Answer 1

a) find the point where the slope is +1: For this, you need to find dy/dx. Do this using implicit differentiation:

x²/25 + y²/4 = 1

find d/dx for each term--treat y as a function of x and therefor use the chain rule for the y term:

2x/25 + 2y/4 * dy/dx = 0. Set dy/dx = 1, and get:

2x/25 = -2y/4

8x = -50y

y = -4x/25

plug this in to the original equation:

x²/25 + (-4x/25)²/4 = 1

x²/25 + 4x²/625 = 1

25x2 + 4x2 = 625

x2 = 625/29

x = ±4.64

b) This ellipse is centered at the origin (no x or y offsets). Therefor, the maximum and minimum x values occur when y = 0

Answer 2

Your question is a little unclear.

The slope of the tangent will be 1 in 2 places, at both of which y=4x/25, in quadrant 1 and quadrant 3.

In order to get the specific co-ordinates, you need to plug the equation above into the equation of the ellipse and solve for x.

x varies from -5 to 5 and y varies from -2 to 2

P.S.

I got the answer for slope by implicit differentiation and then setting the derivative = to 1.