Number of lines which are normal as well as tangents to the curve $y^2=x^3$? Number of lines which are normal as well as tangents to the curve $y^2=x^3$?
The line passes through two points $(x_1,f(x_1))$ and $(x_2,f(x_2))$ on the curve.
What is the general method to solve such problems?I could'nt proceed much.
 A: The set $S:=\{(x,y)\>|\>y^2=x^3\}$ consists of the two arcs
$$\gamma_\pm: \quad y=\pm x^{3/2}\qquad(0\leq x<\infty)\ .$$
A sketch shows that any line tangent to $\gamma_+$ at some point $P$ will intersect $\gamma_-$ at some point $Q$, and one conjectures that there is exactly one point $P=(u^2,u^3)\in\gamma_+$ for which the intersection at $Q$ is orthogonal. The tangent $\tau_P$ at $P$ has slope ${3\over2}(x_P)^{1/2}$, whence the equation
$$\tau_P:\quad y=u^3+{3\over2} u\>(x-u^2)={u\over2}(3x-u^2)\ .\tag{1}$$
Intersecting $\tau_P$ with $S$ leads to the equation
$$x^3=y^2={u^2\over4}(3x-u^2)^2\ .$$
This equation has the obvious double solution $x=u^2$, and the third solution is $x={u^2\over4}$. From $(1)$ we then obtain the point $$Q=\left({u^2\over4},-{u^3\over8}\right)\ .$$
The slope of $\gamma_-$ at $Q$ is $-{3\over2}(x_Q)^{1/2}$, so that the orthogonality condition leads to
$$-{3\over2}\>{u\over2}\cdot {3\over2} u=-1\ .$$
There is a unique positive solution, namely $u={2\over3}\sqrt{2}$, which then determines $P=\bigl({8\over9}, {16\over 27}\sqrt{2}\bigr)$.
It follows that there are exactly two lines satisfying the given condition.
A: Firstly draw a graph of your curve. You'll see there are two distinct sections to it: $\sqrt{x^3}$ and $-\sqrt{x^3}$. For a line to be a tangent and a normal it must be a tangent to one and normal to the other. Also note due to symmetry if we find one then the vertical reflection must also be an answer so lets look at lines which are tangent to $\sqrt{x^3}$ and normal to $-\sqrt{x^3}$.
Differentiating gives:
$$\frac{d}{dx}\left(\pm x^\frac{3}{2}\right)=\pm\frac{3}{2}x^\frac{1}{2}$$
So normal is from a point $x_0$ is: $y=\frac{2}{3\sqrt{x_0}}(x-x_0)-\sqrt{x_0^3}$. This will intersect the positive curve when:
$$\frac{2}{3\sqrt{x_0}}(x-x_0)-\sqrt{x_0^3}=\sqrt{x^3}$$
$$\frac{2(x-x_0)-3x_0^2}{3\sqrt{x_0}}=\sqrt{x^3}$$
$$\frac{4x^2-8x_0x-12x_0^2x+4x_0^2+12x_0^3+9x_0^2}{9x_0}=x^3$$
$$4x^2-8x_0x-12x_0^2x+4x_0^2+12x_0^3+9x_0^2=9x_0x^3$$
$$(x-x_0)(9x_0x^2+9x_0^2x-4x+4x_0+12x_0^2+9x_0^3)=0$$
This will have only two solutions (and hence be a tangent) when the determinant of the quadratic is zero.
$$(9x_0-4)^2-4\cdot9x_0^2\cdot(4x_0+12x_0^2+9x_0^3)=0$$
$$16-216a^2-432a^3-243a^3=0$$
$$-(3x_0+2)^2(9x_0-2)=0$$
As $x_0$ cannot be negative then we have the answer of $x_0=\frac{2}{9}$.
So there are two curves which are both tangent and normal: $y=\pm\left(\frac{2}{3\sqrt{\frac{2}{9}}}(x-\frac{2}{9})-\sqrt{\left(\frac{2}{9}\right)^3}\right)$.
These simplify to: $y=\pm\left(\frac{\sqrt{2}(27x-8)}{27}\right)$.

