A curve C passes through the origin, every point of C has x-coordinate greater than −1, and for every point of C the product of its distance to F(1, 0) and its distance to the line x = a (with a < 0) equals 1. Find a.

Step-by-step solution

1. 1

   Write the two distances algebraically. For a point $P(x,y)$,

   $|PF|=\sqrt{(x-1)^{2}+y^{2}},\qquad \operatorname{dist}(P,\ x=a)=|x-a|$

   The curve lies entirely to the right of the vertical line (its points satisfy $x>-1$ while $a<0$), so $x-a>0$ and the absolute value can be dropped:

   $\sqrt{(x-1)^{2}+y^{2}}\,(x-a)=1$

2. 2

   Use the one concrete point you are given. The condition holds at every point of $C$, and the only point actually named is the origin. Substituting $x=0$, $y=0$ turns the functional condition into a single equation in $a$ — this is the whole idea of the problem.

3. 3

   Substitute the origin.

   $\sqrt{(0-1)^{2}+0^{2}}\cdot(0-a)=1\quad\Longrightarrow\quad 1\cdot(-a)=1$

4. 4

   Solve for a.

   $-a=1\quad\Longrightarrow\quad a=-1$

   This is consistent with the requirement $a<0$ ✓.

   $\boxed{a=-1}$

5. 5

   Check that the sign assumption holds. With $a=-1$ the line is $x=-1$, and the problem states every point of $C$ satisfies $x>-1$, so $x-a=x+1>0$ throughout and dropping the absolute value was legitimate ✓. The curve's equation is therefore

   $\sqrt{(x-1)^{2}+y^{2}}\,(x+1)=1$

   and squaring gives $(x+1)^{2}\left[(x-1)^{2}+y^{2}\right]=1$.

6. 6

   Verify the origin lies on that curve. At $(0,0)$: $(1)^{2}\left[1+0\right]=1$ ✓, so the derived equation really does pass through $O$, confirming the value of $a$.