At which points are the tangents drawn to the ellipse x 2 + y 2 = [ a ] x + [ a ] y parallel to

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

So we have to take the 1st derivative of the equation then equate dy / dx to zero.

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

Therefore the tangents are parallel to the x-axis at points (1, -1.828) and (1, 3.828).

the answer is (-8,-5)

step-by-step explanation:

the endpoint to midpoint is half the length. the length is (-2,-1) so when you are at the midpoint just add it to it to find your answer -6+(-2)= -8 and -4+(-1)=-5 so (-8,-5)

$4034.40

the perimeter of the field is the sum of the lengths of the four sides:

47 yd + 158 yd + 47 yd + 158 yd = 410 yd

the cost of 3 ft (1 yd) of fence is 3×$3.28 = $9.84.

so the cost of jake's wood perimeter fance is

(410 yd) × ($9.84 /yd) = $4034.40