How do we find the latus rectum of parabola when the equation is given in the polar form, $1/r = 1 + \cos t$?

Question

How do we find the latus rectum of parabola when the equation is given in this polar form? $$1/r = 1 + \cos t$$

This curve cuts the $x$ axis on $1/2$ and $y$ axis on $1$ and $-1$.

How can I find the latus rectum of this parabola?

Answer 1

With you points you easily find the equation $y^2=-2x+1$ and so focus lies at $(0,0)$ and latus rectum passes thru $(0.0)$.

Answer 2

Transform the equation $1/r = 1 + \cos t$ to its $xy$-coordinates,

$$1= r + r\cos t=\sqrt{x^2+y^2} + x$$

or, in its canonical form,

$$x= -\frac 12 y^2 + \frac 12$$

For a parabola in the form $x=a(y-h)^2+ k$, its focal point is $(h, k+\frac{1}{4a})$. So, the focus point is (0,0) and its distance to the vertex is $\frac 12$. Thus, the latus rectum $4\times \frac 12 = 2$.