MHB Max and min tangent and triangle

[leprofece]

in what point of the circumference: x² + y² = 1 the tangent to this, (to yhe circunference) form with the coordinate axes the triangle of smaller area?.

answer ( +/-(sqrt2/2), +/-(sqrt2/2) )

Ok y²= 1-x²

Now I don't know in what point must i get the tangent?? I don't think it is in 0,0
I would get y = x

 

[Evgeny.Makarov]

Use the fact the the tangent to the circle $x^2+y^2=1$ at point $(x_0,y_0)$ has equation $xx_0+yy_0=1$.

 

[leprofece]

Use the fact the the tangent to the circle $x^2+y^2=1$ at point $(x_0,y_0)$ has equation $xx_0+yy_0=1$.

ok I thought that but what is x₀ and y₀?

 

[Evgeny.Makarov]

$(x_0,y_0)$ is any point on the circle through which (point) the tangent is drawn.

 

[leprofece]

xx0+yy0=1.
So I must solve for y to introcuced in the another?
y = 1-xx₀/y₀?

Sorry but that way is very dificult to me

 

[Evgeny.Makarov]

y = 1-xx0/y0?

Yes, this is the equation of a tangent, but don't forget what you need to solve. You are interested in the area of the triangle formed by the x- and y-axes and the tangent. The triangle is right-angled, so to find the area, you need to know the sides adjacent to the right angle (also called legs, or catheti). To find the sides, we find the intercept of the tangent, i.e., where the tangent crosses the axes. Let's call the point of tangency $(a,b)$ instead of $(x_0,y_0)$. Substituting $x=0$ into the equation of the tangent
\[
ax+by=1
\]
gives $y=1/b$; similarly, substituting $y=0$ gives $x=1/a$. Thus, twice the area of the triangle is $1/(ab)$. This is the expression we need to minimize subject to the restriction that the point of tangency lies on the circle, i.e., $a^2+b^2=1$. Minimizing $1/(ab)$ is equivalent to maximizing $ab$ or $a^2b^2$ if $a>0$ and $b>0$. Thus, we need to maximize $a^2(1-a^2)$. Denoting $a^2=t$ (note that $a^2$ varies between 0 and 1), we need to maximize $t(1-t)$ on $[0,1]$. The point of maximum is $t=1/2$, from where $a=\sqrt{1/2}$ and $b=\sqrt{1/2}$.

 

[leprofece]

Yes, this is the equation of a tangent, but don't forget what you need to solve. You are interested in the area of the triangle formed by the x- and y-axes and the tangent. The triangle is right-angled, so to find the area, you need to know the sides adjacent to the right angle (also called legs, or catheti). To find the sides, we find the intercept of the tangent, i.e., where the tangent crosses the axes. Let's call the point of tangency $(a,b)$ instead of $(x_0,y_0)$. Substituting $x=0$ into the equation of the tangent
\[
ax+by=1
\]
gives $y=1/b$; similarly, substituting $y=0$ gives $x=1/a$. Thus, twice the area of the triangle is $1/(ab)$. This is the expression we need to minimize subject to the restriction that the point of tangency lies on the circle, i.e., $a^2+b^2=1$. Minimizing $1/(ab)$ is equivalent to maximizing $ab$ or $a^2b^2$ if $a>0$ and $b>0$. Thus, we need to maximize $a^2(1-a^2)$. Denoting $a^2=t$ (note that $a^2$ varies between 0 and 1), we need to maximize $t(1-t)$ on $[0,1]$. The point of maximum is $t=1/2$, from where $a=\sqrt{1/2}$ and $b=\sqrt{1/2}$.

I would really appreciatte it

 

[Evgeny.Makarov]

I would really appreciatte it

But you don't. ;) "Would" is an indicator of a conditional sentence: e.g., "I would go to the cinema if it were not raining". Just kidding. (Smile)