The given equation of the ellipse is x^2
+ y^2 = 2 x + 2 y
At tangent line, the point is horizontal with the x-axis
therefore slope = dy / dx = 0
<span>So we have to take the 1st derivative of the equation
then equate dy / dx to zero.</span>
x^2 + y^2 = 2 x + 2 y
x^2 – 2 x = 2 y – y^2
(2x – 2) dx = (2 – 2y) dy
(2x – 2) / (2 – 2y) = 0
2x – 2 = 0
x = 1
To find for y, we go back to the original equation then substitute
the value of x.
x^2 + y^2 = 2 x + 2 y
1^2 + y^2 = 2 * 1 + 2 y
y^2 – 2y + 1 – 2 = 0
y^2 – 2y – 1 = 0
Finding the roots using the quadratic formula:
y = [-(- 2) ± sqrt ( (-2)^2 – 4*1*-1)] / 2*1
y = 1 ± 2.828
y = -1.828 , 3.828
<span>Therefore the tangents are parallel to the x-axis at points (1, -1.828)
and (1, 3.828).</span>
