Smallest possible area for triangle

Question

I'm solving the following question:

The two legs of a right triangle lie along the positive x and y axes. The hypotenuse is tangent to the ellipse $2x^2 + y^2 = 1$. What is the smallest possible area for such a triangle ?

This is my attempt at solving the problem:

Let $a$ be the length of the x axis and $b$ be the length in it's y axis if the triangle. So it's area is $1/2 a.b$

Let's find the tangent of the ellipse:

$2x^2 + y^2 = 1$

$4x + 2y \dfrac{dy}{dx} = 0$

$\dfrac{dy}{dx} = \dfrac{-2x}{y}$

Since the two points on the triangle are $(0,b)$ and $(a,0)$, we can find the slope of the line:

$m = \dfrac{b-0}{0-a} = \dfrac{-a}{b}$

So the equation of hypotonuse is $y = \dfrac{-a}{b}x + c$

Since $(0,b)$ is one of the point in the line, we can substitute it in the above line to find that $b = c$. Solving it with the other co-ordinate, we can find the following fact $b = a$.

So now area is $1/2 a.b = 1/2 a^2$

Now I used the first derivative test to find that the function attains it's local minima at $0$. So the smallest possible area for such a triangle is zero.

I have missed lots of steps in my above solution for brevity, but I can expand if needed. My question: is zero the right answer ? Unfortunately, the book I'm using doesn't provide the answer to this question.

Answer 1

For $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the equation of the tangent line at $(a\cos t, b \sin t)$ is given by $$\frac{\cos t}ax+\frac{\sin t}by=1$$ which intersects the axes at $\frac a{\cos t}$ and $\frac b{\sin t}$, respectively. Then, the area of the triangle is $$Area = \frac12 \frac a{\cos t}\frac b{\sin t}=\frac{ab }{\sin2t}\ge ab $$ Here, $a=\frac1{\sqrt2}$ and $b=1$, allowing the smallest area $\frac1{\sqrt2}$.

Answer 2

You have made a small error. You wrote that $$m = \dfrac{b-0}{0-a}$$ but then said that this is equal to $$ \dfrac{-a}{b}$$ This is clearly wrong. In fact, $$m=-\frac{b}{a}$$

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The Solution We have $y=-\frac{b}{a}x+b=\frac{ab-bx}{a}$. Substituting this into $2x^2+y^2=1$ yields $$\begin{align} &2x^2+\left(\frac{ab-bx}{a}\right)^2=1\\ \implies& 2a^2x^2+b^2x^2-2ab^2x+a^2b^2=a^2\\ \implies& x^2(2a^2+b^2)+x(-2ab^2)+(a^2b^2-a^2)=0\\ \end{align}$$ We could then go on to try to solve this quadratic, but in fact we can do better. Note that the tangent line will intersect the ellipse once, and only once. Hence, the discriminant of this quadratic in $x$ is equal to $0$: $$4a^2b^4-4(2a^2+b^2)(a^2b^2-a^2)=0$$ We can divide through by $a^2$, as we don't have a degenerate triangle with $a=0$, simplifying to $$\begin{align}&4b^4-4(2a^2+b^2)(b^2-1)=0\\ \implies&4b^4-4(2a^2b^2-2a^2+b^4-b^2)=0\\ \implies & a^2(8b^2-8)=4b^2\\ \implies & a^2=\frac{b^2}{2(b^2-1)} \end{align}$$ Hence, we have, assuming without loss of generality that $a,b\geqslant 0$, that $$a=\frac{b}{\sqrt{2(b^2-1)}}$$ So we find that the area of the triangle, which we know is equal to $\frac{1}{2}ab$, is in fact equal to $$\frac{b^2}{2\sqrt{2(b^2-1)}}$$ So we now have an expression for the area, $A$, in terms of one variable only! This means that to find the minimum value of $A$, we can just differentiate with respect to $b$, solve for stationary points, check it is a minimum, and we are basically done!

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I hope that was helpful. Thank you for giving me the opportunity to attack this interesting question; I'd never come across a similar one before. What I find particularly pleasing about this solution is that we have no need to look at the derivative of the ellipse, which would've made things rather complicated. If you have any questions please don't hesitate to ask!

Answer 3

Outline : We will write down the equation of tangent at arbitrary point $P$ of ellipse. Let it intersect the coordinate axes in $A$ and $B$. We identify the legs of the right triangle formed can be identified as $x$- and $y$- intercepts of the tangent line. So we find the smallest area of triangle minimizing the product of the intercepts.

Let $P=(h,k)$. Then tangent at $P$ is $2xh+ky=1$ which can be written as $$\frac{x}{1/(2h)}+\frac{y}{1/k}=1$$

Area $=\frac{1}{2}\cdot \frac{1}{(2h)}\cdot \frac{1}{k}\cdot$ where $2h^2+k^2=1$

By AM-GM, $$\frac{2h^2+k^2}{2} \ge \sqrt{2h^2\cdot k^2}$$ $$\Rightarrow \frac{1}{hk} \ge \frac{2\sqrt{2}}{2h^2+k^2}=2\sqrt{2}$$ $$\therefore A_{\text{min}}=\frac{1}{4hk}=\frac{2\sqrt{2}}{4}=\frac{1}{\sqrt{2}}$$

Equality occurs at $2h^2=k^2$ which can be used to determine coordinates of optimal $P$.

Answer 4

If we consider just the part of the ellipse in the first quadrant, it is the graph of the function $f(x) = \sqrt{1 - 2x^2}$ on the interval $\left[0,\frac1{\sqrt2}\right]$; and between $0$ and $\frac1{\sqrt2}$, $f$ is a positive function with a decreasing first derivative.

From my answer to The method to calculate area between f(x) and coordinate axes we therefore know that the area between the coordinate axes and the line tangent to the graph of $f$ is minimized when the product $x f(x)$ is maximized, and that area is $2 x_m f(x_m)$ where $x_m$ maximizes the product.

In this case, $x f(x) = x \sqrt{1 - 2x^2}.$ It is simpler to maximize $x^2 (f(x))^2,$ which is $x^2 (1 - 2x^2),$ which is $u(1-2u)$ where $u = x^2.$ The derivative of $u(1-2u)$ is $1 - 4u,$ which is zero at $u = \frac14,$ positive for $u < \frac14$, and negative for $u > \frac14,$ so the maximum occurs at $u = \frac14$, therefore at $x = \frac12,$ and the area of the triangle is

$$ 2 \left(\frac12\right) \sqrt{1 - 2\left(\frac12\right)^2} = \sqrt{\frac12}. $$

Answer 5

Let $(h,k)$ be the point on ellipse from where you draw the tangent, then, $$2h^2+k^2=1$$

You already know the slope of tangent is $$\frac{dy}{dx}=\frac{-2h}{k}$$

Constructing the equation of hypotenuse from this,

$$\frac{y-k}{x-h}=\frac{-2h}{k}$$

Putting $y=0$ , you get $x=a$,

$$a=\frac{1}{2h}$$ Similarly $$b=\frac{1}{k}$$

You need to minimise the area, i.e. , $$=\frac{ab}{2}$$

$$=\frac{1}{4hk}$$

$$=\frac{1}{4\cdot{k}\cdot{}\sqrt{\frac{1-k^2}{2}}}$$

$$=\frac{1}{2\sqrt{2}\cdot{k}\cdot{}\sqrt{1-k^2}}$$ Can you continue?