How to determine the angle between the tangents to $4y^2=x^2(2-x^2)$ at the origin?

Question

I received this questions to solve but I hook at a certain stage.

The equation of a curve is $4y²=x²(2-x²)$. Determine the equation of the tangent at the origin and show that the angle between these tangents is $\tan^{-1}(2\sqrt{2})$.

• Now, I find the derivative using implicit differentiation and I got $dy/dx=x(1-x)/2y$

• after finding that, I input the values of $x$ and $y$ at the origin but the answer I got is indeterminate. I feel unsatisfied with that Thus, since the gradient is indeterminate. How I can show that an angle exist. Pls help

Answer 1

You may proceed several ways. Here is one approach.

First note that your curve is symmetrical about both the $x$ and $y$ axes.

Solving for the curve's components in the upper half-plane (quadrants I and II), we get

$$y=\frac{1}{2}\sqrt{x^2(2-x^2)}$$

Differentiating gives $y'=\frac{1}{4\sqrt{x^2(2-x^2)}}(x^2(-2x)+2x(2-x^2))$, i.e.

$$y'=\frac{x(1-x^2)}{\sqrt{x^2(2-x^2)}}$$

Now when $x\to0^{+}$ we see $y'\to\frac{1}{\sqrt{2}}$, while when $x\to0^{-}$ we see $y'\to-\frac{1}{\sqrt{2}}$. (Here I use the fact that $\sqrt{x^2}=|x|$, which is $x$ when $x>0$ and $-x$ when $x<0$.)

By the curve's symmetry, we conclude the "tangents" at the origin have slopes $\pm\frac{1}{\sqrt{2}}$. I assume you can take it from here.

---

Added. Here's another approach, using your strategy of implicit differentiation.

Differentiating implicitly gives

$$8yy'=x^2(-2x)+2x(2-x^2)\implies y'=\frac{4x-4x^3}{8y}=\frac{x(1-x^\color{red}{2})}{2y}$$

So you're missing an exponent of $2$.

Now just consider the limit of $y'$ as we approach the origin along the curve $y=\frac{1}{2}\sqrt{x^2(2-x^2)}$:

$$\lim_{x\to0}\frac{x(1-x)}{\sqrt{x^2(2-x^2)}}$$

This gives the same result above: the left-sided limit is $-\frac{1}{\sqrt{2}}$ and the right-sided limit is $\frac{1}{\sqrt{2}}$.

Answer 2

The tangent cone at the origin is the lowest degree terms of the equation $$4y^2-x^2(2-x^2)=0,$$ i.e. $$4y^2-2x^2=0,$$ or $$y=\pm \frac{x}{\sqrt{2}}.$$

Now the angle between these two lines is twice the angle of the slope $2\tan^{-1}(\frac1{\sqrt{2}}),$ then use the identity $2\tan^{-1}(t)=\tan^{-1}(\frac{2t}{1-t^2}).$

Answer 3

Here's a solution with trigonometry, parameterization, and conic sections:

We have $4y^2=x^2(2-x^2)=2x^2-x^4$ .

So we also have $x^4-2x^2+4y^2=0$

Completing the square: $(x^2-1)^2+4y^2=1$

Let $u=x^2-1$ and $v=2y$, then we have $u^2+v^2=1$, the equation of a circle. This suggests the parameterizations, $x^2=1+\cos{t}$ and $y=\frac{1}{2}\sin{t}$

By the half-angle trig identity, we can simplify the $x^2$ term, $x^2=2\cos^2{t/2}$.

Then take the square root of both sides, $x=\sqrt{2}\cos{t/2}$ and we keep in mind the $\sqrt{2}$ term can be positive or negative.

We want our $x$ and $y$ to have functions with the same argument, so we now let $y=\sin{t/2}\cos{t/2}$.

We want the slope of the tangent line at the origin, that occurs when $x=y=0$. In turn, this occurs when t is an odd multiple of $\pi$, $t=(2k+1)\pi$ for integer $k$.

The slope of the tangent line at (x,y) is given by $\frac{dy/dt}{dx/dt}$

$\frac{dy}{dt}=\frac{1}{2}(\cos^2{t/2}-\sin^2{t/2})$

and $\frac{dx}{dt}=\frac{-\sqrt{2}\sin{t/2}}{2}$

So $\frac{dy}{dx}=(-\frac{\sqrt{2}}{2})\frac{\cos^2{t/2}-\sin^2{t/2}}{\sin{t/2}}$

We want $t$ to be odd multiples of $\pi$ from the above arguments. At these values, the numerator of the non-constant term becomes $-1$ and the denominator is $+1$ or $-1$. Keeping in mind that $\sqrt{2}$ can be replaced with $-\sqrt{2}$, we have

$t=(2k+1)\pi: \frac{dy}{dx}=(+/-)\frac{\sqrt{2}}{2}$

And now you can use the Inverse Tangent formula Michael Ubmande lists above.

Parameterization can often help disambiguate limits and derivatives and actually serves to define differentiation in more advanced math.

Answer 4

A change of coordinates leads to a short cut with no (explicit) calculus. If we use polar coordinates: $$x=r\cos{\theta}$$

$$y=r\sin{\theta}$$

Swapping in for $4y^2=x^2(2-4x^2)$:

$$4r^2\sin^2{\theta}=r^2\cos^2{\theta}(2-4r^2\cos^2{\theta})$$

Note the $r^2$ term cancels except the one in the parenthetical term. After some algebra we get:

$$4\tan^2{\theta}=2-r^2\cos^2{\theta}$$ Rearranging and swapping in a trig identity,

$$r^2=(2-4\tan^2{\theta})(1+\tan^2{\theta})$$

$r^2$ then becomes zero when $\tan^2{\theta}=\frac{1}{2}$ and we can now solve for $\tan{\theta}$ and and again use the Inverse Tangent Identity.