Elliptic Curve?

Relevant wiki: First Order Differential Equations - Problem Solving

I will solve the problem more generally, and derive an equation for any curve that is perpendicular to the $E_r$ at all its points.

Let $(x,y)$ be a point on $E_r$ . Then the tangent to $E_r$ at this point satisfies $2x\:dx + 2ay\:dy = 0\ \ \ \therefore\ \ \ \left.\frac{dy}{dx}\right|_{E_r} = -\frac{x}{ay}.$ If a curve $K$ is perpendicular to $E_r$ , then it must satisfy $\left.\frac{dy}{dx}\right|_{E_r}\cdot \left.\frac{dy}{dx}\right|_K = -1\ \ \ \therefore\ \ \ \left.\frac{dy}{dx}\right|_K = \frac{ay}{x}.$ This is a differential equation for $K$ . We solve it by separating the variables, then integrating: $\frac 1 a \frac{dy}{y} = \frac{dx}{x}\ \ \ \therefore\ \ \ \frac 1 a\ln {|y|} = \ln{|x|} + c \ \ \ \therefore\ \ \ y = C\:x^a.$ (In fact, we can "glue" together any solutions for $x < 0$ and for $x > 0$ .)

Clearly, if $y = x^3$ is one of these curves, we must have $a = \boxed{3}$ .

I don't know if this is correct, but I share. Hope I explain properly, I don't speak good english

Let be $f(x,y)= x^2+ay^2$ and $g(x,y) = x^3-y$

Also, let $L_h(c) = \{ x \in \mathbb{R}^n : h(x) = c \}$ for some function $h$ . Notice that $L_f(r^2)$ are our elipses and $L_g(0)$ our curve.

Suppose $L_f(r^2)$ is perpendicular to $L_g(0)$ for all $r \in S$ Then the tangent lines to those curves from the intersection point $(x_y,y_0)$ are perpendicular.

Be have that if $(x_0,y_0) \in L_f(r^2)$ then $\bigtriangledown f(x_0,y_0)$ is perpendicular to the tangent from $(x_0, y_0)$ of $L_f(r^2)$

But then $\bigtriangledown f(x_0,y_0)$ is parallel to the tangent from $(x_0, y_0)$ of $L_g(0)$ . We know that the tangent vector from $(x_0,y_0 )$ to $\ L_g(0)$ is $(1, 3x_0^2)$ .

So, we finally have (using $y_0 = x_0^3$ ):

$\bigtriangledown f(x_0,y_0) = \lambda (1, 3 x_0^2)$

$(2x_0, 2ax_0^3)= \lambda (1, 3 x_0^2)$

$2x_0 (1, ax_0^2) = \lambda (1,3x_0^2)$

And is clearly $\lambda = 2x_0$ which gives us $\boxed{ a = 3}$

$E_r : x^2 + ay^2 = r^2, r > 0$ and $y = x^3$ intersect perpendicularly.

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Let's find the derivatives.

${E_r}' : 2x + 2ayy' = 0 \Rightarrow y' = -\frac{x}{ay}$ and $y' = 3x^2$ .

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They intersect perpendicularly, and therefore there must exist an infinite number of $x$ and $y$ for which $y = \frac{3}{a}x^3$ since there is an infinite number of possible values for $r$ and because no pair of ellipses in the family can intersect each other.

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However, we already know that $y = x^3$ , so there must be an infinite number of $x$ and $y$ for which $x^3 = \frac{3}{a}x^3$ holds. But now we notice that if $a \neq 3$ , the equation has a finite number of solutions. Hence, $a = \boxed{3}$ .