Question

The locus of the mid point of the focal radii of a variable point moving on the parabola, 𝑦2 = 4𝑎𝑥 is a parabola whose

• A. latus rectum is half that of the original parabola
• B. directix is the y-axis
• C. focus has the co-ordinates (a,0)
• D. vertex has the co-ordinates (a/2,0)

Answer 1

Answer: (A), (B), (C), (D)

All of the above

Answer 2

To find the locus of the midpoint of the focal radii of a variable point moving on the parabola $y^2 = 4ax$, we follow these steps:

1. Identify the focus and a general point on the parabola: The given parabola is $y^2 = 4ax$. Its focus is $S(a, 0)$. Let $P(x_1, y_1)$ be a variable point on the parabola. We can parameterize this point as $P(at^2, 2at)$.

2. Define the midpoint: Let $M(h, k)$ be the midpoint of the focal radius $SP$. Using the midpoint formula for $S(a, 0)$ and $P(at^2, 2at)$: $h = \frac{a + at^2}{2}$ $k = \frac{0 + 2at}{2}$

3. Express $h$ and $k$ in terms of the parameter $t$: $h = \frac{a(1 + t^2)}{2} \quad \cdots (1)$ $k = at \quad \cdots (2)$

4. Eliminate the parameter $t$ to find the locus: From equation (2), we can express $t$ as $t = \frac{k}{a}$. Substitute this value of $t$ into equation (1): $h = \frac{a(1 + (\frac{k}{a})^2)}{2}$ $h = \frac{a(1 + \frac{k^2}{a^2})}{2}$ $h = \frac{a + \frac{k^2}{a}}{2}$ Multiply by 2: $2h = a + \frac{k^2}{a}$ Multiply by $a$: $2ah = a^2 + k^2$ Rearrange the terms to get the equation in standard form: $k^2 = 2ah - a^2$ $k^2 = 2a(h - \frac{a}{2})$

5. Replace $(h, k)$ with $(x, y)$ to get the equation of the locus: The locus of the midpoint is $y^2 = 2a(x - \frac{a}{2})$.

6. Analyze the properties of the resulting parabola: This equation is of the form $Y^2 = 4AX$, where $Y = y$, $X = x - \frac{a}{2}$, and $4A = 2a \Rightarrow A = \frac{a}{2}$.

   • Vertex: The vertex is at $(X=0, Y=0)$, which means $x - \frac{a}{2} = 0 \Rightarrow x = \frac{a}{2}$ and $y = 0$. So, the vertex is $(\frac{a}{2}, 0)$.

   • Focus: The focus is at $(X=A, Y=0)$, which means $x - \frac{a}{2} = \frac{a}{2} \Rightarrow x = a$ and $y = 0$. So, the focus is $(a, 0)$. This is the same as the focus of the original parabola $y^2 = 4ax$.

   • Directrix: The directrix is $X = -A$, which means $x - \frac{a}{2} = -\frac{a}{2} \Rightarrow x = 0$. So, the directrix is $x=0$ (the y-axis).

   • Length of Latus Rectum: The length of the latus rectum is $4A = 4(\frac{a}{2}) = 2a$. This is half the length of the latus rectum of the original parabola ($4a$).

The locus of the midpoint of the focal radii of a variable point moving on the parabola, $y^2 = 4ax$ is a parabola whose:

• Equation is $y^2 = 2a(x - \frac{a}{2})$
• Vertex is $(\frac{a}{2}, 0)$
• Focus is $(a, 0)$ (which is the same as the focus of the original parabola)
• Directrix is $x=0$ (the y-axis)
• Length of latus rectum is $2a$