<s>[INST] 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.
 [/INST] 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: $y=\pm\left(\frac{\sqrt{2}(27x-8)}{27}\right)$.

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