Find the length of normal chord which subtends an angle of `90^0` at the vertex of the parabola `y^2=4xdot`

Options

A. `6sqrt3`
B. `3sqrt3`
C. `2`
D. `1`

Answer

To find the length of the normal chord that subtends an angle of \(90^\circ\) at the vertex of the parabola \(y^2 = 4x\), we can follow these steps:

### Step 1: Identify the parameters of the parabola
The given parabola is \(y^2 = 4x\). This can be compared with the standard form \(y^2 = 4ax\), where \(a = 1\).

### Step 2: Set up the normal chord
Let the endpoints of the normal chord be represented by the parameters \(t_1\) and \(t_2\). The coordinates of the points on the parabola corresponding to these parameters are:
- Point A: \((t_1^2, 2at_1) = (t_1^2, 2t_1)\)
- Point B: \((t_2^2, 2at_2) = (t_2^2, 2t_2)\)

### Step 3: Use the properties of the normal
The equation of the normal at point A is given by:
\[ y - 2t_1 = -\frac{1}{t_1}(x - t_1^2) \]
This can be rearranged to find the relationship between \(t_1\) and \(t_2\):
\[ t_2 = -\frac{4}{t_1} \]

### Step 4: Use the angle condition
Since the normal chord subtends an angle of \(90^\circ\) at the vertex, we have the condition:
\[ t_1 \cdot t_2 = -4 \]
Substituting \(t_2 = -\frac{4}{t_1}\) into this condition gives:
\[ t_1 \left(-\frac{4}{t_1}\right) = -4 \]
This confirms the relationship.

### Step 5: Substitute and solve for \(t_1\)
From the equation \(t_2 = -\frac{4}{t_1}\), we can substitute into the normal equation:
\[ -\frac{4}{t_1} = -t_1 - \frac{2}{t_1} \]
Multiplying through by \(t_1\) (assuming \(t_1 \neq 0\)):
\[ -4 = -t_1^2 - 2 \]
Rearranging gives:
\[ t_1^2 = 2 \quad \Rightarrow \quad t_1 = \sqrt{2} \]

### Step 6: Find \(t_2\)
Now substituting \(t_1\) back to find \(t_2\):
\[ t_2 = -\frac{4}{\sqrt{2}} = -2\sqrt{2} \]

### Step 7: Calculate the coordinates of points A and B
Using the values of \(t_1\) and \(t_2\):
- Point A: \((t_1^2, 2t_1) = (2, 2\sqrt{2})\)
- Point B: \((t_2^2, 2t_2) = (8, -4\sqrt{2})\)

### Step 8: Calculate the length of the chord
Using the distance formula:
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ AB = \sqrt{(8 - 2)^2 + (-4\sqrt{2} - 2\sqrt{2})^2} \]
\[ = \sqrt{6^2 + (-6\sqrt{2})^2} \]
\[ = \sqrt{36 + 72} = \sqrt{108} = 6\sqrt{3} \]

### Conclusion
The length of the normal chord that subtends an angle of \(90^\circ\) at the vertex of the parabola \(y^2 = 4x\) is \(6\sqrt{3}\).

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