If two points $A(-2, a)$ and $B(4, \beta)$ are such that the triangle $AOB$ is the right-angle triangle right angle at $O$. If $S = 0$ be the equation of the locus of the foot of the perpendicular $P$ drawn from the point $O$ from the line $AB$, then:
Step-by-Step Solution
Key Concept: A variable family of circles passing through two fixed curves always passes through their points of intersection and one additional fixed point.
Step 1: Identify the general form of the locus $S=0$.
The provided solution defines the equation of the locus $S=0$ by substituting a relation into a circle equation, resulting in:
$$S \equiv \left(x^2 + y^2 - \frac{1}{n}y\right) - a\left(x - \frac{m}{n}y\right) = 0$$
Expanding and rearranging this equation, we get:
$$x^2 + y^2 - ax - \frac{1}{n}y + \frac{am}{n}y = 0$$
$$x^2 + y^2 - ax + \left(\frac{am-1}{n}\right)y = 0$$
This equation is of the general form $Ax^2 + Ay^2 + Bx + Cy + D = 0$. Since the coefficients of $x^2$ and $y^2$ are equal (both 1) and there is no $xy$ term, this equation represents a circle for any real values of $m, n, a$ (provided $n \neq 0$).
Step 2: Interpret the given equation as a family of curves.
The equation $S \equiv (x^2 + y^2 - \frac{1}{n}y) - a(x - \frac{m}{n}y) = 0$ is in the form $S_1 + \lambda L_1 = 0$, where $S_1 \equiv x^2 + y^2 - \frac{1}{n}y = 0$ represents a circle, $L_1 \equiv x - \frac{m}{n}y = 0$ represents a line, and $\lambda = -a$ is a parameter. This form represents a family of curves (specifically, circles in this case) that all pass through the intersection points of the circle $S_1=0$ and the line $L_1=0$. These intersection points are the fixed points common to all circles in the family, irrespective of the parameter $a$.
Step 3: Determine the fixed points through which the family of circles passes.
To find the points of intersection of $S_1=0$ and $L_1=0$, we solve them simultaneously:
The line equation is $x - \frac{m}{n}y = 0 \implies x = \frac{m}{n}y$.
Substitute this into the circle equation $x^2 + y^2 - \frac{1}{n}y = 0$:
$$\left(\frac{m}{n}y\right)^2 + y^2 - \frac{1}{n}y = 0$$
$$\frac{m^2}{n^2}y^2 + y^2 - \frac{1}{n}y = 0$$
$$y^2\left(\frac{m^2}{n^2} + 1\right) - \frac{1}{n}y = 0$$
$$y^2\left(\frac{m^2 + n^2}{n^2}\right) - \frac{1}{n}y = 0$$
Factor out $y$:
$$y\left(y\frac{m^2 + n^2}{n^2} - \frac{1}{n}\right) = 0$$
This gives two possible values for $y$:
1. $y = 0$
2. $y\frac{m^2 + n^2}{n^2} - \frac{1}{n} = 0 \implies y\frac{m^2 + n^2}{n^2} = \frac{1}{n} \implies y = \frac{n^2}{n(m^2 + n^2)} = \frac{n}{m^2 + n^2}$
Now, find the corresponding $x$ values using $x = \frac{m}{n}y$:
1. If $y=0$, then $x = \frac{m}{n}(0) = 0$. So, the first fixed point is $(0,0)$.
2. If $y = \frac{n}{m^2 + n^2}$, then $x = \frac{m}{n}\left(\frac{n}{m^2 + n^2}\right) = \frac{m}{m^2 + n^2}$. So, the second fixed point is $\left(\frac{m}{m^2+n^2}, \frac{n}{m^2+n^2}\right)$.
These two points are the fixed points through which all circles in the family $S=0$ pass, as stated in the original solution.
Step 4: Evaluate the given options.
Based on the analysis of the locus $S=0$:
* **Option 3: Locus is a parabola.** From Step 1, the equation $S=0$ represents a circle, not a parabola. Thus, this option is incorrect.
* **Option 4: Locus is a circle.** From Step 1, the equation $S=0$ is indeed the general equation of a circle. Thus, this option is correct.
* **Option 1: $(1, 3)$ lies on $S=0$.** For a point to lie on $S=0$ (i.e., on all circles in the family $S=0$) irrespective of the parameter $a$, it must be one of the fixed points identified in Step 3. Let's check if $(1,3)$ can be the second fixed point $\left(\frac{m}{m^2+n^2}, \frac{n}{m^2+n^2}\right)$.
If $\frac{m}{m^2+n^2} = 1$ and $\frac{n}{m^2+n^2} = 3$.
From these, we have $m = m^2+n^2$ and $n = 3(m^2+n^2)$.
Substitute $m^2+n^2 = n/3$ (from the second equation) into the first equation:
$m = \frac{n}{3}$.
Now, substitute $m = n/3$ back into the expression for $m^2+n^2$:
$m^2+n^2 = \left(\frac{n}{3}\right)^2 + n^2 = \frac{n^2}{9} + n^2 = \frac{10n^2}{9}$.
Equating the two expressions for $m^2+n^2$:
$$\frac{10n^2}{9} = \frac{n}{3}$$
Assuming $n \neq 0$ (as $n=0$ would lead to $m=0$, and the fixed point being $(0,0)$), we can divide by $n$:
$$\frac{10n}{9} = \frac{1}{3} \implies n = \frac{9}{30} = \frac{3}{10}$$
Then, $m = \frac{n}{3} = \frac{1}{3} \cdot \frac{3}{10} = \frac{1}{10}$.
Since we found consistent values for $m$ and $n$ (specifically $m=1/10$ and $n=3/10$) for which $(1,3)$ is one of the fixed points, this means $(1,3)$ can lie on $S=0$ for all values of $a$. Thus, this option is also correct.
The final answer is $\boxed{\text{1,4}}$.
Correct Answer: 1,4