In parabola y^2=4x, From the point (15,12), three normals are drawn then centroid of triangle formed by three Co normal points is

Options

A. (16/3.0)
B. (4,0)
C. (26/3.0)
D. (6,0)

Explanation

To solve the problem of finding the centroid of the triangle formed by the three normal points drawn from the point (15, 12) to the parabola \(y^2 = 4x\), we can follow these steps:

### Step 1: Identify the Parabola
The given parabola is \(y^2 = 4x\). From this, we can identify that \(a = 1\) (since \(4a = 4\)).

### Step 2: Write the Equation of the Normal
The equation of the normal to the parabola at the point corresponding to the parameter \(t\) is given by:
\[ y + tx = 2a(t^2 + a) \]
Substituting \(a = 1\), we have:
\[ y + tx = 2(t^2 + 1) \]
This simplifies to:
\[ y + tx = 2t^2 + 2 \]

### Step 3: Substitute the Point (15, 12)
Since the normal passes through the point (15, 12), we substitute \(x = 15\) and \(y = 12\) into the normal equation:
\[ 12 + 15t = 2t^2 + 2 \]
Rearranging gives us:
\[ 2t^2 - 15t - 10 = 0 \]

### Step 4: Solve the Quadratic Equation
We can solve the quadratic equation \(2t^2 - 15t - 10 = 0\) using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 2\), \(b = -15\), and \(c = -10\):
\[ t = \frac{15 \pm \sqrt{(-15)^2 - 4 \cdot 2 \cdot (-10)}}{2 \cdot 2} \]
Calculating the discriminant:
\[ t = \frac{15 \pm \sqrt{225 + 80}}{4} = \frac{15 \pm \sqrt{305}}{4} \]

### Step 5: Find the Roots
Let \(t_1 = \frac{15 + \sqrt{305}}{4}\), \(t_2 = \frac{15 - \sqrt{305}}{4}\), and we can find \(t_3\) using the property that the sum of the roots \(t_1 + t_2 + t_3 = 0\). Thus:
\[ t_3 = -\left(t_1 + t_2\right) = -\left(\frac{15 + \sqrt{305}}{4} + \frac{15 - \sqrt{305}}{4}\right) = -\frac{30}{4} = -\frac{15}{2} \]

### Step 6: Find the Coordinates of the Normal Points
The coordinates of the points on the parabola corresponding to \(t_1\), \(t_2\), and \(t_3\) are:
- For \(t_1\): \((t_1^2, 2t_1)\)
- For \(t_2\): \((t_2^2, 2t_2)\)
- For \(t_3\): \((t_3^2, 2t_3)\)

### Step 7: Calculate the Centroid
The centroid \(G\) of the triangle formed by these points is given by:
\[ G = \left(\frac{t_1^2 + t_2^2 + t_3^2}{3}, \frac{2t_1 + 2t_2 + 2t_3}{3}\right) \]
Since \(2t_1 + 2t_2 + 2t_3 = 0\), the y-coordinate of the centroid is \(0\).

Now, we need to calculate \(t_1^2 + t_2^2 + t_3^2\):
Using the identity \(t_1^2 + t_2^2 + t_3^2 = (t_1 + t_2 + t_3)^2 - 2(t_1t_2 + t_2t_3 + t_3t_1)\):
- We already know \(t_1 + t_2 + t_3 = 0\).
- The product \(t_1 t_2 t_3 = -\frac{10}{2} = -5\) (from the original quadratic).
- The sum of the products \(t_1 t_2 + t_2 t_3 + t_3 t_1 = -\frac{15}{2}\).

Thus:
\[ t_1^2 + t_2^2 + t_3^2 = 0 - 2(-5) = 10 \]

Finally, the x-coordinate of the centroid is:
\[ G_x = \frac{10}{3} \]

### Final Answer
The coordinates of the centroid \(G\) are:
\[ G = \left(\frac{10}{3}, 0\right) \]