L (1, 3) and `L^(1) = (1, -1)` are the ends of latus rectum of a parabola, then area of quadrilateral formed by tangents and normals at L and `L^(1)` ( in Square Units ) is

Options

A. 2
B. 4
C. 8
D. 16

Answer

To solve the problem, we need to find the area of the quadrilateral formed by the tangents and normals at the points L(1, 3) and L'(1, -1), which are the ends of the latus rectum of a parabola. Here is the step-by-step solution:

### Step 1: Identify the parabola
The ends of the latus rectum are given as L(1, 3) and L'(1, -1). The latus rectum of a parabola is a line segment perpendicular to the axis of symmetry and passes through the focus. The standard form of a parabola that opens horizontally is given by \( y^2 = 4ax \).

### Step 2: Find the value of 'a'
The length of the latus rectum is given by \( 4a \). To find 'a', we first calculate the distance between the points L and L':
\[ \text{Distance} = \sqrt{(1 - 1)^2 + (3 - (-1))^2} = \sqrt{0 + (3 + 1)^2} = \sqrt{16} = 4 \]
Thus, the length of the latus rectum is 4, which gives us:
\[ 4a = 4 \implies a = 1 \]

### Step 3: Write the equation of the parabola
Substituting \( a = 1 \) into the standard form, we get:
\[ y^2 = 4(1)x \implies y^2 = 4x \]

### Step 4: Find the equations of the tangents at points L and L'
Using the formula for the tangent to the parabola \( y^2 = 4ax \) at the point \( (x_1, y_1) \):
\[ yy_1 = 2a(x + x_1) \]
For point L(1, 3):
\[ y(3) = 2(1)(x + 1) \implies 3y = 2x + 2 \implies 2x - 3y + 2 = 0 \]

For point L'(1, -1):
\[ y(-1) = 2(1)(x + 1) \implies -y = 2x + 2 \implies 2x + y + 2 = 0 \]

### Step 5: Find the equations of the normals at points L and L'
The equation of the normal to the parabola at point \( (x_1, y_1) \) is given by:
\[ y - y_1 = -\frac{1}{2a}(x - x_1) \]
For point L(1, 3):
\[ y - 3 = -\frac{1}{2}(x - 1) \implies y = -\frac{1}{2}x + \frac{1}{2} + 3 \implies y = -\frac{1}{2}x + \frac{7}{2} \]

For point L'(1, -1):
\[ y + 1 = -\frac{1}{2}(x - 1) \implies y = -\frac{1}{2}x + \frac{1}{2} - 1 \implies y = -\frac{1}{2}x - \frac{1}{2} \]

### Step 6: Find the points where tangents and normals intersect the x-axis
For the tangent at L:
Setting \( y = 0 \):
\[ 2x - 3(0) + 2 = 0 \implies 2x + 2 = 0 \implies x = -1 \]
Thus, point T is (-1, 0).

For the tangent at L':
Setting \( y = 0 \):
\[ 2x + 0 + 2 = 0 \implies 2x + 2 = 0 \implies x = -1 \]
Thus, point N is also (-1, 0).

### Step 7: Calculate the area of the quadrilateral
The area of the quadrilateral formed by points T, L, L', and N can be calculated as the sum of the areas of triangles TLL' and NLL'.

#### Area of Triangle TLL':
Using the formula for the area of a triangle given vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\):
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting points T(-1, 0), L(1, 3), L'(1, -1):
\[ \text{Area} = \frac{1}{2} \left| -1(3 - (-1)) + 1(-1 - 0) + 1(0 - 3) \right| \]
\[ = \frac{1}{2} \left| -1(4) - 1 + (-3) \right| = \frac{1}{2} \left| -4 - 1 - 3 \right| = \frac{1}{2} \left| -8 \right| = 4 \]

#### Area of Triangle NLL':
Using the same formula with points N(-1, 0), L(1, 3), L'(1, -1):
\[ \text{Area} = \frac{1}{2} \left| -1(3 - (-1)) + 1(-1 - 0) + 1(0 - 3) \right| = 4 \]

### Step 8: Total Area of Quadrilateral
Thus, the total area of the quadrilateral formed by tangents and normals is:
\[ \text{Total Area} = \text{Area of TLL'} + \text{Area of NLL'} = 4 + 4 = 8 \text{ square units} \]

### Final Answer
The area of the quadrilateral formed by the tangents and normals at points L and L' is **8 square units**.