Question

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

Answer 1

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

Answer:

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)

Answer 2

Answer:

point B because the figure flips over the line of reflection, the rightmost point of the reflected figure will be the leftmost point of the original object

Step-by-step explanation: