Determine the point on the curve $a ^ 2 x ^ 2 + y ^ 2 = a ^ 2$ in the first quadrant such that the area of ​the triangle by tangent

Question

Problem: Let the $a$ arbitrary . Determine the point on the curve $a ^ 2 x ^ 2 + y ^ 2 = a ^ 2$ in the first quadrant such that the area of ​​the triangle by tangent the curve drawn at this point closes the axes is the lowest possible .
My work:
Let $y=kx+l$ be that tangent line in the first quadrant. We know that line goes through points $(0,l)$ and $(-\frac{l}{k},0).$ So area is $A=\frac{l*-\frac{l}{k}}{2}=-\frac{-l^2}{2k}$ But area depends of two variables and we want to depends on just one because we will than solving equation: $f'(l)$ or $f'(k)$.
Ok let's do this. First of all we will need $y'$ and we will find it by implicit differentiation. $$a ^ 2 x ^ 2 + y ^ 2 = a ^ 2$$ $$2a^2x + 2yy'= 0$$ $$a^2x + yy'= 0$$ $$ yy'= -a^2x $$ $$ y'= \frac{-a^2x}{y} $$ So we know that $k=\frac{-a^2x}{y}$. By using equation of tangent line we can use write $l$ as variable of parametar $x$ and $y$. $$y=kx+l$$ $$y-kx=l$$ $$y-\frac{-a^2x^2}{y}=l$$ $$\frac{y^2-a^2x^2}{y}=l$$ Ok now we have again that our formula for area depends of two variables. But let's use fact that $(x,y)$ is in the first quadrant. $$a ^ 2 x ^ 2 + y ^ 2 = a ^ 2$$ $$y^2=a^2-a^2x$$ $$y=(a^2-a^2x)^{(\frac{-1}{2})}$$ We don't have negative solution because dot is in the first quadrant.
Finally now we have want we want. Plugin in want we know in area formula and search solution for $A'(x)=0$ Then we check second derivative to be sure it is our solution max or min. And we can easily chech what $a$ is.
Is this ok or maybe you have better idea?

Answer 1

You can parameterize this ellipse with $x=\cos\theta$ and $y=a\sin\theta$ and obtain the equation of the tangent at $\theta$ as $$ax\cos\theta+y\sin\theta=a$$

Then the area of the triangle is $$\frac 12\sec\theta \operatorname{cosec}\theta=\frac{a}{\sin 2\theta}$$ and this has minimum value $a$. This occurs when $\theta=\frac{\pi}{4}$ thus giving the coordinates $(x,y)$