Consider the ellipse plotted by x^2+9y^2=9. Find the y-coordinate of the point on the ellipse that is farthest from the point (0,1).
Your answer should be expressed as a fraction of whole numbers.
Let the point be (h,k)
The distance d is thus
d^2 = x^2 + (y-1)^2
but, we know that x^2 = 9 - 9y^2, so
d^2 = 9-9y^2 + (y^2-2y+1)
= 10 - 2y - 8y^2
dd/dy = -2 - 16y = -2(1+8y)
so, y = -1/8
and you can get x
To find the y-coordinate of the point on the ellipse that is farthest from the point (0,1), we need to use the concept of calculus and optimization.
Step 1: We can start by rearranging the given equation of the ellipse to isolate y^2:
x^2 + 9y^2 = 9
9y^2 = 9 - x^2
y^2 = (9 - x^2)/9
Step 2: Now, we need to find the derivative of y^2 with respect to x in order to find the slope of the tangent line at any point on the ellipse.
Taking the derivative of y^2 with respect to x gives:
d/dx (y^2) = d/dx ((9 - x^2)/9)
=> 2y * dy/dx = (-2x)/9
=> dy/dx = (-2x)/(9 * 2y)
=> dy/dx = -x/(9y)
Step 3: Next, we can set up the equation of the line passing through the point (0,1) and having the slope dy/dx:
(y - 1) = (-x/(9y))(x - 0)
=> y - 1 = -x^2/(9y)
Step 4: Simplify the equation further by cross-multiplying:
y^2 - y = -x^2/9
Step 5: Substitute for y^2 from the ellipse equation:
(9 - x^2)/9 - y = -x^2/9
9 - x^2 - 9y = -x^2
9y = 9
y = 9/9
y = 1
So the y-coordinate of the point on the ellipse that is farthest from the point (0,1) is 1.