Question

If \( P \) is a point which divides the line segment joining the focus of the parabola \( y^2 = 12x \) and a point on the parabola in the ratio 1:2, then the locus of \( P \) is:

• A. \( y^2 = 2(x - 2) \)
• B. \( y^2 = 4x \)
• C. \( y^2 = 4(x - 2) \)
• D. \( y^2 = 9(x - 3) \)

Hint:

When finding the locus of a point dividing a segment in a given ratio, use the section formula to calculate the coordinates of the point and then form the equation of the locus.

Answer 1

The Correct Option is C

To solve the problem, we'll follow these steps: Step 1: Identify the focus of the parabola The given parabola is \( y^2 = 12x \). The standard form of a parabola \( y^2 = 4ax \) has its focus at \( (a, 0) \). Comparing with \( y^2 = 12x \), we get: \[ 4a = 12 \implies a = 3 \] Thus, the focus of the parabola is \( (3, 0) \). Step 2: Parametrize a point on the parabola A general point on the parabola \( y^2 = 12x \) can be represented as \( (t^2, 2\sqrt{3}t) \), where \( t \) is a parameter. This satisfies the equation: \[ (2\sqrt{3}t)^2 = 12(t^2)
12t^2 = 12t^2 \] Step 3: Find the coordinates of \( P \) The point \( P \) divides the line segment joining the focus \( (3, 0) \) and the point on the parabola \( (t^2, 2\sqrt{3}t) \) in the ratio \( 1:2 \). Using the section formula, the coordinates of \( P \) are: \[ P = \left( \frac{2 \cdot 3 + 1 \cdot t^2}{1 + 2}, \frac{2 \cdot 0 + 1 \cdot 2\sqrt{3}t}{1 + 2} \right)
P = \left( \frac{6 + t^2}{3}, \frac{2\sqrt{3}t}{3} \right)
P = \left( 2 + \frac{t^2}{3}, \frac{2\sqrt{3}t}{3} \right) \] Let \( P = (x, y) \). Then: \[ x = 2 + \frac{t^2}{3} \quad \text{(1)}
y = \frac{2\sqrt{3}t}{3} \quad \text{(2)} \] Step 4: Eliminate the parameter \( t \) From equation (2), solve for \( t \): \[ y = \frac{2\sqrt{3}t}{3}
t = \frac{3y}{2\sqrt{3}}
t = \frac{\sqrt{3}y}{2} \] Substitute \( t = \frac{\sqrt{3}y}{2} \) into equation (1): \[ x = 2 + \frac{\left( \frac{\sqrt{3}y}{2} \right)^2}{3}
x = 2 + \frac{\frac{3y^2}{4}}{3}
x = 2 + \frac{y^2}{4} \] Rearrange: \[ x
- 2 = \frac{y^2}{4}
y^2 = 4(x
- 2) \] Final Answer: The locus of \( P \) is \( y^2 = 4(x
- 2) \). \[ \boxed{y^2 = 4(x
- 2)} \]