2
$\begingroup$

A variable parabola touches the $x$-axis and $y$-axis at $A(1,0)$ and $B(0,1)$ on the co-ordinate plane respectively. Now, we are required to find the locus of the focus of this variable parabola.

The process to arrive at this locus is a standard one, and goes like,

Starting with facts (observations),

  1. The parabola has $x$- and $y$-axes as its tangents, and it lies in the first quadrant
  2. We know that these tangents intersect orthogonally and hence the intersection point lies on its directrix.
  3. Since the directrix passes through the origin let its equation be $y=mx$.
  4. Now $A(1,0)$ and $B(0,1)$ lies on the parabola hence if we define focus as $F(h,k)$ we find that from the definition of parabola $$\begin{align} FA &= \text{(distance from $A$ to the directrix)} \\ FB &=\text{(distance from $B$ to the directrix)} \end{align}$$ Hence we have sufficient conditions to get the locus,

Writing, $$ (FA)^2 = (h-1)^2 + (k-0)^2 = \frac{|(0)-m(1)|^2}{1+m^2}$$

$$ (FB)^2 = (h-0)^2 + (k-1)^2 = \frac{|(1)-m(0)|^2}{1+m^2}$$

Adding both and simplifying we get the locus of $F(h,k)$ as,

$$x^2 + y^2 - x - y + 0.5 = 0$$

This is an imaginary equation which doesn't give the locus of $F(h,k)$, So my question is how to interpret this result, What does it mean to have a set of imaginary focal points? or Are there is any reason to claim that my solution process is wrong? If yes, then what is the correct way to obtain the locus of $F(h,k)$?

$\endgroup$
0
1
$\begingroup$

Contrary to what you wrote, $x=y=\frac12$ satisfies your last equation: the circle isn’t imaginary but consists of a single point. This is as it should be: two points and the tangents at those points uniquely determine a parabola.

I’ve seen a version of this problem in which the parabola is tangent to the coordinate axes, but at a pair of unspecified points. Perhaps that’s what was intended here.

$\endgroup$
4
  • $\begingroup$ but how did you figure out the only real solution of that equation? is there any standard way of doing it? any article on those imaginary equations would be helpful.. $\endgroup$ – user774954 Apr 18 '20 at 8:40
  • 1
    $\begingroup$ @MukunthA.G The equation is that of a circle. If you substitute the coordinates of its center into the left-hand side, the resulting value is the negative of the square of its radius. (Try this with the generic circle equation for yourself!) When I did that, I got $0$, so the equation actually represents a single point—the center of the degenerate circle. $\endgroup$ – amd Apr 18 '20 at 8:44
  • 1
    $\begingroup$ @MukunthA.G Complete the square. $$(x^2-x+1/4) + (y^2-y+1/4) = (x-1/2)^2 + (y-1/2)^2 = 0.$$ Therefore, $x = y = 1/2$ is a solution. $\endgroup$ – heropup Apr 18 '20 at 9:13
  • $\begingroup$ @heropup Right. After completing the squares of $x^2+y^2-2hx-2ky+f=0$ you have $(x-h)^2+(y-k)^2+f'=0$, so the left-over constant term $f'$ is clearly equal to the value of the original expression at $x=h$, $y=k$. $\endgroup$ – amd Apr 18 '20 at 20:10
0
$\begingroup$

OP has a fixed parabola $$\sqrt{x}+\sqrt{y}=1,$$ it have only a fixed pint, so no locus.

To have family of tncated parabolas touching $x$ and $y$ axes at $(c,0)$ and $(0,c)$ The equation of the parabola is $$\sqrt{x}+\sqrt{y}=\sqrt{c},~~~~(1)$$ We can rationalize (1) to have full parabola as: $$\left(\frac{x-y}{\sqrt{2}}\right)^2=c\sqrt{2}\frac{(x+y+c/2)}{\sqrt{2}} \implies L_1^2=4AL_2,~~ L_1~\text{perpendicular to}~ L_2$$ The Eq. of axis of the parabola is $L_1= 0 \implies y=x$ and Equation of the latus rectum is $L_2=A$, their intersection gives the focus $F$ as $$y=x, L_2=\frac{x+y-c/2}{\sqrt{2}}=\frac{c}{2\sqrt{2}} \implies F~is~ (c/2,c/2)$$ Therefor the locus of the focus of the family of parablolas (1) is $y=x$ which is the fixed axis of the parabolas,

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy