How long is the tangent that intercepts between the coordinate axes?

[leprofece]

348) given the length of the ellipse to a² h² + b²y² = a² b² Find the length of the tangent shorter, that intercepts between?
the coordinate axes

answer L = a + b

The equations a² h² + b²y² = a² b²

And the line y = mx + b as the objetive function

 

[MarkFL]

This is what I am guessing is being asked (formulated in more standard notation):

Given the ellipse:

\(\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

and a line tangent to the ellipse, find the shortest possible distance along such a line between its two coordinate axes intercepts.

Because of symmetry of the ellipse, centered at the origin, in the 4 quadrants of the plane, I would work in the first quadrant where everything is positive.

Let's begin by finding the family of tangent lines. We may represent this line with:

\(\displaystyle y=mx+c\)

So, let's substitute for $y$ into the equation of the ellipse:

\(\displaystyle \frac{x^2}{a^2}+\frac{(mx+c)^2}{b^2}=1\)

Multiply through by $a^2b^2$:

\(\displaystyle b^2x^2+a^2(mx+c)^2=a^2b^2\)

Expand binomial and distribute:

\(\displaystyle b^2x^2+a^2m^2x^2+2a^2cmx+a^2c^2=a^2b^2\)

Arrange in standard quadratic form in $x$:

\(\displaystyle \left(b^2+a^2m^2\right)x^2+2a^2cmx+a^2\left(c^2-b^2\right)=0\)

Now, since the line is tangent to the ellipse, the discriminant must be zero:

\(\displaystyle 4a^4c^2m^2-4\left(b^2+a^2m^2\right)a^2\left(c^2-b^2\right)=0\)

What do you get for $m$?

Once you find $m$, you then need to find the intercepts of the resulting line, and set as your objective function the distance between the line's intercepts (use the square of the distance for simplicity), minimize this function, take the square root, and the result you posted will follow.

 

[leprofece]

Ohh again I tried to get m and I got somethimg very long which was verified by me in wolfrang alpha

I got m = b/a(sqrt ((b²-c²)/(2c²- b²)

and the linl in wolfang alpha is
http://www.wolframalpha.com/input/?i=a^2m^2c^2+%3D%28b^2%2Ba^2m^2%29%28b^2-c^2%29

 

[MarkFL]

That's not what I got for $m$. And without actually seeing your work, I have no idea where you went wrong. The first thing I did was divide through by $4a^2$ to get:

\(\displaystyle a^2c^2m^2-\left(b^2+a^2m^2\right)\left(c^2-b^2\right)=0\)

Now distribute:

\(\displaystyle a^2c^2m^2-\left(b^2c^2-b^4+a^2c^2m^2-a^2b^2m^2\right)=0\)

\(\displaystyle -b^2c^2+b^4+a^2b^2m^2=0\)

Divide through by $b^2$:

\(\displaystyle -c^2+b^2+a^2m^2=0\)

\(\displaystyle m^2=\frac{c^2-b^2}{a^2}\)

Take the negative root since the slope of the tangent line must be negative:

\(\displaystyle m=-\frac{\sqrt{c^2-b^2}}{a}\)

Thus, the tangent line is:

\(\displaystyle y=-\frac{\sqrt{c^2-b^2}}{a}x+c\)

So, what are the two intercepts, and the square of the distance between them?

 

[Opalg]

Here is another approach to this problem (no better or worse than MarkFL's method – both approaches involve a fair bit of calculation although the final result is surprisingly simple).

My method relies on knowing the parametric form of the ellipse equation $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$, in which the general point on the curve is given by $(a\cos\theta,b\sin\theta).$ The gradient at that point is $-\dfrac{b\cos\theta}{a\sin\theta}$, so the equation of the tangent is $y - b\sin\theta = -\dfrac{b\cos\theta}{a\sin\theta}(x - a\cos\theta).$ If you put $y=0$ in that equation then you see that the tangent meets the $x$-axis when $x = a\sec\theta.$ Similarly, the tangent meets the $y$-axis when $y = b\csc\theta$. So if $d$ is the length of the intercept then $d^2 = a^2\sec^2\theta + b^2\csc^2\theta.$ Differentiate that expression to find that the minimum occurs when $\tan^2\theta = \dfrac ba$ (I am leaving out all the details of the calculations). Finally, put that value into the formula for $d^2$ and you find that $d^2 = (a+b)^2.$ So when you take the square root, you get a really simple answer!

 

[leprofece]

That's not what I got for $m$. And without actually seeing your work, I have no idea where you went wrong. The first thing I did was divide through by $4a^2$ to get:

\(\displaystyle a^2c^2m^2-\left(b^2+a^2m^2\right)\left(c^2-b^2\right)=0\)

Now distribute:

\(\displaystyle a^2c^2m^2-\left(b^2c^2-b^4+a^2c^2m^2-a^2b^2m^2\right)=0\)

\(\displaystyle -b^2c^2+b^4+a^2b^2m^2=0\)

Divide through by $b^2$:

\(\displaystyle -c^2+b^2+a^2m^2=0\)

\(\displaystyle m^2=\frac{c^2-b^2}{a^2}\)

Take the negative root since the slope of the tangent line must be negative:

\(\displaystyle m=-\frac{\sqrt{c^2-b^2}}{a}\)

Thus, the tangent line is:

\(\displaystyle y=-\frac{\sqrt{c^2-b^2}}{a}x+c\)

So, what are the two intercepts, and the square of the distance between them?

OK ABOUT THE OTHER METHOD
The problem is that students has not studied parametrics yet
so I must take your method
So Intercepts are
y = c
and
x = -(ac)/((c²-b²)^1/2

the square of the distance between them
(c-0)²+ (-(ac)/((c²-b²)^1/2)²

that is = c²+ (a²c²)/(c² -b²)

Factor comun c²(a²/c²-b²)

Now how can I proceed ??
minimize this function, take the square root, and the result you posted will follow.

Ok minimize function but in terms of a or b or c ?

 

[MarkFL]

...Ok minimize function but in terms of a or b or c ?

$a$ and $b$ are fixed, it is $c$ that will vary.

By the way, although it does not impact your result, the $x$ intercept of the line is positive, not negative. And so your objective function is:

\(\displaystyle f(c)=c^2+\frac{a^2c^2}{c^2-b^2}\)

I would leave it in this form to make the differentiation somewhat simpler. Your factorization was incorrect anyway.

 

[leprofece]

Ok Derivating I got
2c +2a²c(c²-b²)-2c(a²c²/( c²-b²)²

symplifying and after some algebra I got 2c (1 - a²b²) = 0
and I got a =1/b and b = 1/a
And Then ?

 

[MarkFL]

Okay, if I squint my eyes and ignore that you have not properly used bracketing symbols, I agree that:

\(\displaystyle f'(c)=2c+\frac{\left(c^2-b^2\right)2a^2c-a^2c^2(2c)}{\left(c^2-b^2\right)^2}=0\)

Now, given that $c\ne0$, we may divide through by $2c$ to obtain:

\(\displaystyle 1+\frac{\left(c^2-b^2\right)a^2-a^2c^2}{\left(c^2-b^2\right)^2}=0\)

Next, let's multiply through by \(\displaystyle \left(c^2-b^2\right)^2\). We know that if $c=b$ then the $x$-intercept would be at infinity, and so that is certainly not the minimum we are after, and so carrying this out we now obtain:

\(\displaystyle \left(c^2-b^2\right)^2+\left(c^2-b^2\right)a^2-a^2c^2=0\)

If we distribute in the middle term, we get:

\(\displaystyle \left(c^2-b^2\right)^2+a^2c^2-a^2b^2-a^2c^2=0\)

Combine like terms...

\(\displaystyle \left(c^2-b^2\right)^2-a^2b^2=0\)

Now, factor as the difference of squares:

\(\displaystyle \left(c^2-b^2+ab\right)\left(c^2-b^2-ab\right)=0\)

Now, from this we find the following implication:

\(\displaystyle c^2=b^2\pm ab\)

Now, we know that $b^2<c^2$ and so we must have:

\(\displaystyle c^2=b^2+ab\)

Now, going back to our objective function, and substituting for $c^2$, we obtain:

\(\displaystyle f=b^2+ab+\frac{a^2\left(b^2+ab\right)}{b^2+ab-b^2}\)

Factor:

\(\displaystyle f=\left(b^2+ab\right)\left(1+\frac{a^2}{ab}\right)\)

\(\displaystyle f=b(a+b)\left(\frac{a+b}{b}\right)=(a+b)^2\)

Now, since $f$ is the square of the distance $D$ we are minimizing, we then conclude:

\(\displaystyle D_{\min}=\sqrt{f}=a+b\)