# A property of the focus of the cissoid of Diocles

I would like to prove the following result:

Let $\mathcal C$ be the cissoid of Diocles having equation $x(x^2 + y^2) = b y^2$, where $b > 0$. Let $O$ be its singular point and let $F$ be its focus$^*$. Then, for every point $P$ of $\mathcal C$, the following equality holds: $$(FP - OP)(FP + 3 OP) = OF^2.$$

I already know that $O = (0, 0)$ and $F = (4b, 0)$. Since $\mathcal C$ can be parameterized by $$P = \left ( \frac{b t^2}{1 + t^2}, \frac{b t^3}{1 + t^2} \right ) \qquad (t \in \mathbb R)$$ one could prove the result by computing the lengths of all the segments involved. Indeed, $$FP = \frac b {\sqrt{1 + t^2}} (4 + t^2) \qquad OP = \frac b {\sqrt{1 + t^2}} t^2 \qquad OF = 4b$$ from which it is easy to see that $$(FP - OP)(FP + 3 OP) = 16b^2 \qquad OF^2 = 16 b^2.$$

But this proof gives no insight on why the result should be true. So what I am looking for is a geometric proof of the result.

Thank you in advance for any contribution.

${}^*$ Recall that a non-singular point $F$ is said to be a focus of a curve $\mathcal C$ if two tangent lines having slopes $i$ and $-i$ can be drawn from $F$ to $\mathcal C$.

We can use the fact that the cissoid $\mathcal C$ is the inverse curve of the parabola $\mathcal C'$ having equation $x = b y^2$ with respect to the unit circle centered at its vertex $O$. Then, the focus $F$ of $\mathcal C$ is the inverse of the focus $F'$ of $\mathcal C'$, as shown in the following figure:
Here $P$ is a point of the cissoid $\mathcal C$ and $P'$ its inverse on the parabola $\mathcal C'$.
By definition of inversion we have: $$OP \cdot OP' = 1 \qquad OF \cdot OF' = 1$$ and so by the secant-secant power theorem the points $P, P', F, F'$ belong to the same circle. Hence, $F'\widehat F P = P\widehat {P'} F'$ because they are angles subtended by the same arc $PF'$ at the circumference. Then also $O\widehat F P = O \widehat {P'} F'$, and of course $P\widehat O F = P'\widehat O F'$, from which it follows that the triangles $OFP$ and $OP'F'$ are similar. Therefore: $$\frac {FP} {F'P'} = \frac {OF} {OP'}.$$
Now we can write the equality in terms of $OP'$, $OF'$ and $F'P'$. First, $$OP = \frac 1 {OP'} \qquad OF = \frac 1 {OF'} \qquad FP = \frac{F'P' \cdot OF} {OP'} = \frac {F'P'} {OP' \cdot OF'}$$ then the equality becomes: $$\left ( \frac {F'P'} {OP' \cdot OF'} - \frac 1 {OP'} \right ) \left ( \frac {F'P'} {OP' \cdot OF'} + 3 \cdot \frac 1 {OP'} \right ) = \frac 1 {OF'^2}$$ which reduces to: $$(F'P' - OF')(F'P' + 3 OF') = OP'^2.$$