# Thread: Hypocycloid. Oh so fun!

1. ## Hypocycloid. Oh so fun!

At the risk of asking too many questions on my first day.

I need some more help.

How do I "Show that every tangent line to the curve x^(2/3) +y^(2/3)=1 in the first quadrant has the property that the portion of the line in the first quadrant has length 1" using implicit differentiation?

Thanks bunches for any help.

2. Hello, Columbo!

Show that every tangent line to the curve $x^{\frac{2}{3}} +y^{\frac{2}{3}} \;=\;1$
in the first quadrant has the property that the portion of the line
in the first quadrant has length 1, using implicit differentiation.
Find the derivative . . .

. . $\frac{2}{3}x^{-\frac{1}{3}} + \frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx} \;=\;1 \quad\Rightarrow\quad \frac{dy}{dx} \;=\;-\frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}}}$

Let $A(p,q)$ be a point on the astroid in Quadrant 1.
. . Note that: . $p^{\frac{2}{3}} + q^{\frac{2}{3}} \:=\:1$

The slope of the tangent at $A$ is: . $m \;=\;-\frac{q^{\frac{1}{3}}}{p^{\frac{1}{3}}}$

The equation of the line through $A$ with slope $m$ is:

. . $y -q \;=\;-\frac{q^{\frac{1}{3}}}{p^{\frac{1}{3}}}(x - p) \quad\Rightarrow\quad y \;=\;-\frac{q^{\frac{1}{3}}}{o^{\frac{1}{3}}}x + p^{\frac{2}{3}}q^{\frac{1}{3}} + q$

. . $y \;=\;-\frac{q^{\frac{1}{3}}}{p^{\frac{1}{3}}}x + q^{\frac{1}{3}}\underbrace{\left(p^{\frac{2}{3}} + q^{\frac{2}{3}}\right)}_{\text{This is 1}}$ . $\Rightarrow\quad\boxed{ y \;=\;-\frac{q^{\frac{1}{3}}}{p^{\frac{1}{3}}}x + q^{\frac{1}{3}}}$

This tangent has: .x-intercept $\left(p^{\frac{1}{3}},\:0\right)$ . . . and y-intercept $\left(0,\:q^{\frac{1}{3}}\right)$

The length of that segment is: . $L \;=\;\sqrt{\left(p^{\frac{1}{3}}-0\right)^2 + \left(0-q^{\frac{1}{3}}\right)^2}$

. . $\text{and we have: }\;L \;=\;\sqrt{\underbrace{p^{\frac{2}{3}} + q^{\frac{2}{3}}}_{\text{This is 1}}} \;=\;\sqrt{1} \;=\;1 \quad\hdots\quad \text{ There!}$