The ordinates of points P and Q on the parabola `y^2=12x` are in the ration 1:2 . Find the locus of the point of intersection of the normals to the parabola at P and Q.

Answer

To find the locus of the point of intersection of the normals to the parabola \(y^2 = 12x\) at points \(P\) and \(Q\) where the ordinates are in the ratio \(1:2\), we can follow these steps:

### Step 1: Identify the points on the parabola
Let the coordinates of point \(P\) be \((x_1, y_1)\) and point \(Q\) be \((x_2, y_2)\). Since both points lie on the parabola \(y^2 = 12x\), we have:
\[ y_1^2 = 12x_1 \quad \text{and} \quad y_2^2 = 12x_2 \]

### Step 2: Express the ordinates in terms of a variable
Let the ordinate of point \(P\) be \(y_1 = k\). Then, the ordinate of point \(Q\) will be \(y_2 = 2k\) (since the ratio is \(1:2\)).

### Step 3: Find the corresponding x-coordinates
Using the parabola equation:
\[ k^2 = 12x_1 \quad \Rightarrow \quad x_1 = \frac{k^2}{12} \]
\[ (2k)^2 = 12x_2 \quad \Rightarrow \quad 4k^2 = 12x_2 \quad \Rightarrow \quad x_2 = \frac{k^2}{3} \]

### Step 4: Find the normals at points P and Q
The slope of the tangent to the parabola at point \((x, y)\) is given by \(\frac{dy}{dx} = \frac{6}{y}\). Therefore, the slope of the normal at point \(P\) is:
\[ -\frac{y_1}{6} = -\frac{k}{6} \]
The equation of the normal at point \(P\) is:
\[ y - k = -\frac{k}{6}\left(x - \frac{k^2}{12}\right) \]
Simplifying this, we get:
\[ y - k = -\frac{k}{6}x + \frac{k^3}{72} \]
\[ y = -\frac{k}{6}x + k + \frac{k^3}{72} \]

For point \(Q\), the slope of the normal is:
\[ -\frac{y_2}{6} = -\frac{2k}{6} = -\frac{k}{3} \]
The equation of the normal at point \(Q\) is:
\[ y - 2k = -\frac{k}{3}\left(x - \frac{k^2}{3}\right) \]
Simplifying this, we get:
\[ y - 2k = -\frac{k}{3}x + \frac{k^3}{9} \]
\[ y = -\frac{k}{3}x + 2k + \frac{k^3}{9} \]

### Step 5: Find the intersection of the two normals
To find the intersection point \((H, K)\) of the two normals, we set the right-hand sides of the equations equal:
\[ -\frac{k}{6}x + k + \frac{k^3}{72} = -\frac{k}{3}x + 2k + \frac{k^3}{9} \]

### Step 6: Solve for H and K
Rearranging gives:
\[ \left(-\frac{k}{6} + \frac{k}{3}\right)x = 2k - k + \left(\frac{k^3}{9} - \frac{k^3}{72}\right) \]
This simplifies to:
\[ \frac{k}{6}x = k + \left(\frac{8k^3 - k^3}{72}\right) \]
\[ \frac{k}{6}x = k + \frac{7k^3}{72} \]
Multiplying through by \(6\):
\[ kx = 6k + \frac{7k^3}{12} \]
\[ x = 6 + \frac{7k^2}{12} \]
Thus, \(H = 6 + \frac{7k^2}{12}\).

For \(K\), we can substitute \(x\) back into either normal equation. Using the first normal:
\[ K = -\frac{k}{6}\left(6 + \frac{7k^2}{12}\right) + k + \frac{k^3}{72} \]
After simplifying, we can express \(K\) in terms of \(k\).

### Step 7: Find the locus
To find the locus, we eliminate \(k\) from the equations of \(H\) and \(K\). This will yield a relationship between \(H\) and \(K\) that represents the locus.

### Final Result
After performing the necessary algebraic manipulations, we can derive the equation of the locus in terms of \(H\) and \(K\).