Find the point at which the rightmost vertical tangent line touches the ellipse. The rightmost vertical tangent line touches the
2 answers:
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)
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