Maximizing the area of the triangle with vertices $(2,0)$, $(0,1)$, $(a,b)$ (with $a>0$ and $b>0$) inscribed in ellipse $x^2 + 4y^2 =4$

Question

I am stuck on a math problem. The problem is

On the ellipse $x^2 + 4y^2 =4$ a point is selected $(a,b)$ in the first quadrant. Draw the ellipse in the $xy$-plane. Also draw a triangle with corners in $(2,0)$, $(0,1)$, and $(a,b)$. The triangle's area is $\frac12a + b-1$.

How should the point $(a,b)$ be selected to reach maximum area of the triangle? Also, motivate why the triangle's area is $\frac12a + b-1$.

I have both drawn the ellipse and the triangle. And now I don't know the next step.

I am not either sure that the drawing is right. It would be much appreciated if I could get any tips to help me solve this problem.

Answer 1

From the base of the triangle, we have to find the point which, if we let a line parallel to the base pass through that point, that line is the farthest away from the base. Obviously, the farthest line is tangent to the ellipse and parallel to the base. The equation of the base is $y=-\frac{1}{2}x+1$, so any parallel line will have the form $y=-\frac{1}{2}x+q$. If we transform the ellipse function into a function and consider the positive side only, we get $y=\sqrt{1-\frac{x^2}{4}}$. Let the two be equal and we get a 2nd degree polynomial in $x$ (and $q$). Solving for $x$, the root's argument in the solution, in order to let the line be a tangent, must be equal to $0$ (the solution for $x$ must be unique). Under the root there is $2-q^2$ and so $q=\sqrt{2}$ since, by looking at the graph, we are only interested in the positive values of $q$. The line we are interested in is then $y=-\frac{1}{2}x+\sqrt2$, and the point is $(\sqrt2,\frac{\sqrt2}{2})$. The distance between two lines of the form $y=mx+q_{1/2}$ is $$d=\frac{\left|q_1-q_2\right|}{\sqrt{m^2+1}}$$. Plugging in the numbers we get $d=\frac{-2+2\sqrt2}{\sqrt5}$ which is the height of the triangle, and the base is $\sqrt5$. Use the triangle area formula and get $-1+\sqrt2$, which is coherent with the formula. The same reasoning can be applied for any $(a,b)$, changing $q$ accordingly to the new coordinates.

Answer 2

The equation of the line passing through $(0,1)$ and $(2,0)$ is: $$y=-\frac12x+1 \Rightarrow x+2y-2=0.$$

The distance from the point $(a,b)$ to the line $x+2y-2=0$ is: $$d=\frac{|a+2b-2|}{\sqrt{1^2+2^2}}=\frac{a+2b-2}{\sqrt5}$$

The area of the triangle is: $$S=\frac12\cdot \sqrt{(2-0)^2+(0-1)^2}\cdot \frac{a+2b-2}{\sqrt5}=\frac a2+b-1.$$

Hint: The maximum area of triangle occurs when the height of the triangle is maximum. Hence, the point $(a,b)$ must be at the center of the arc. Can you show these and find the point $(a,b)$.

Answer 3

Let the given major and minor vertices be $A(2, 0)$ and $B(0, 1)$. The midpoint of $AB$ is $M(1, \frac {1}{2})$. The center of the ellipse is $O(0, 0)$.

Diameter $OM$ meets the curve in quadrant I at point $C(a, b)$. The tangent line at $C$ is parallel to $AB$. Therefore $C$ is the point in that segment furthest from chord $AB$.

Line $OM$: $x = 2y$

$a = 2b$

$x^2 + 4y^2 = 4$

$(2b)^2 + 4b^2 = 4$

$b^2 = \frac{1}{2}$

$b = \frac {\sqrt{2}} {2}$ ... (positive because $C$ is in quadrant I)

$a = 2b = \sqrt{2}$

$C(\sqrt{2}, \frac{\sqrt {2}}{2})$

Why the area formula?

Line $AB: x + 2y - 2 = 0$

distance from $C$ to line $AB = \frac {|a + 2b - 2|} {\sqrt{1^2 + 2^2}} = \frac {|a + 2b - 2|} {\sqrt{5}}$

$AB = \sqrt{5}$

area($\triangle ABC$) = $\frac {1}{2}(\sqrt{5})(\frac {|a + 2b - 2|} {\sqrt{5}}) = \frac{a}{2} + b - 1$ ... (because $a + 2b - 2 > 0$)