Find the point at which the rightmost vertical tangent line touches the ellipse. The rightmost vertical tangent line touches the ellipse at the point

The missing part of the question showing the figure and other earlier questions has been attached.

A) dy/dx = (2x + y)/(-x - 2y)

B) the lower horizontal tangent line is defined by the equation: y = -⅔√15

C) the right most tangent is; x = ⅔√15

D) point at which the rightmost vertical tangent line touches the Ellipse is y = -⅓√15.

Tangent point is at: (⅔√15, -⅓√15)

Step-by-step explanation:

We are given the equation as:

x² + xy + y² = 5. Looking at this equation and the image given, we can see that it is an Ellipse.

A) Using implicit differentiation, we have;

2x(dx) + y(dx) + x(dy) + 2y(dy) = 0

Collecting like terms and rearranging gives us;

(2x + y)dx = (-x - 2y)dy

Thus;

dy/dx = (2x + y)/(-x - 2y)

b) The horizontal tangent will be the point where dy/dx = 0

Thus;

(2x + y)/(-x - 2y) = 0

Therefore,

2x + y = 0

y = -2x

Now, since the horizontal tangent has y to be a constant, then x = -y/2

Putting -y/2 for x in the initial Ellipse equation gives;

(-y/2)² + (-y/2)y + y² = 5

¼y² - ½y² + y² = 5

¾y² = 5

y² = 20/3

y = √(20/3)

y = ± ⅔√15

Therefore,the lower horizontal tangent line is defined by the equation: y = -⅔√15

c) The vertical tangent is gotten by equating the denominator of dy/dx to 0

Thus;

(-x - 2y) = 0

Or when rearranging gives; x + 2y = 0

From symmetry, we can say that the solution is x = ± ⅔√15, with the right most tangent being x = ⅔√15

d) The point at which the right most tangent touches the Ellipse is at;

y = -x/2

Since x = ⅔√15 from C above, then;

y = -(⅔√15)/2

y = -⅓√15

Therefore,the tangent point is at; (⅔√15, -⅓√15)