P is a point on the parabola `y^(2)=4ax` whose ordinate is equal to its abscissa and PQ is focal chord, R and S are the feet of the perpendiculars from P and Q respectively on the tangent at the vertex, T is the foot of the perpendicular from Q to PR, area of the triangle PTQ is

Options

A. `75a^(2)//4`
B. `85a^(2)//2`
C. `75a^(2)//8`
D. `45a^(2)//2`

Answer

To solve the problem step by step, we will follow the given information and derive the area of triangle PTQ.

### Step 1: Identify the coordinates of point P
Given that P is a point on the parabola \( y^2 = 4ax \) and the ordinate (y-coordinate) is equal to the abscissa (x-coordinate), we can denote the coordinates of point P as:
\[ P(t_1) = (at_1^2, 2at_1) \]
Since the ordinate equals the abscissa, we have:
\[ 2at_1 = at_1^2 \implies t_1^2 - 2t_1 = 0 \implies t_1(t_1 - 2) = 0 \]
Thus, \( t_1 = 0 \) or \( t_1 = 2 \). We will consider \( t_1 = 2 \) for point P:
\[ P = (4a, 4a) \]

### Step 2: Determine the coordinates of point Q
Since PQ is a focal chord, we know that if \( t_1 \) is the parameter for point P, then the parameter for point Q, \( t_2 \), is given by:
\[ t_1 \cdot t_2 = -1 \implies t_2 = -\frac{1}{t_1} = -\frac{1}{2} \]
Thus, the coordinates of point Q are:
\[ Q(t_2) = \left( a\left(-\frac{1}{2}\right)^2, 2a\left(-\frac{1}{2}\right) \right) = \left( \frac{a}{4}, -a \right) \]

### Step 3: Find the equations of the tangent at the vertex
The equation of the tangent to the parabola at the vertex (0,0) is:
\[ y = 0 \]
This line is the x-axis.

### Step 4: Find the feet of the perpendiculars R and S
The feet of the perpendiculars from points P and Q to the tangent at the vertex (x-axis) will have the same x-coordinates as P and Q, and their y-coordinates will be 0:
\[ R = (4a, 0), \quad S = \left( \frac{a}{4}, 0 \right) \]

### Step 5: Determine the coordinates of point T
Point T is the foot of the perpendicular from Q to line PR. The slope of line PR can be calculated as follows:
\[ \text{slope of PR} = \frac{0 - 4a}{4a - \frac{a}{4}} = \frac{-4a}{\frac{15a}{4}} = -\frac{16}{15} \]
The slope of the perpendicular line from Q to PR will be the negative reciprocal:
\[ \text{slope of QT} = \frac{15}{16} \]
Using point-slope form from point Q:
\[ y + a = \frac{15}{16}\left(x - \frac{a}{4}\right) \]
Setting \( y = 0 \) to find T:
\[ 0 + a = \frac{15}{16}\left(x - \frac{a}{4}\right) \]
Solving for x gives:
\[ \frac{15}{16}x - \frac{15a}{64} = -a \implies \frac{15}{16}x = -a + \frac{15a}{64} = -\frac{64a}{64} + \frac{15a}{64} = -\frac{49a}{64} \]
Thus:
\[ x = -\frac{49a}{64} \cdot \frac{16}{15} = -\frac{49a}{60} \]
So, T is:
\[ T = \left(-\frac{49a}{60}, 0\right) \]

### Step 6: Calculate the area of triangle PTQ
The area of triangle PTQ can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right| \]
Substituting the coordinates:
\[ P(4a, 4a), Q\left(\frac{a}{4}, -a\right), T\left(-\frac{49a}{60}, 0\right) \]
The area becomes:
\[ \text{Area} = \frac{1}{2} \left| 4a\left(-a - 0\right) + \frac{a}{4}\left(0 - 4a\right) + \left(-\frac{49a}{60}\right)\left(4a - (-a)\right) \right| \]
Calculating each term:
\[ = \frac{1}{2} \left| -4a^2 + \frac{-4a^2}{4} - \frac{49a}{60}(5a) \right| \]
\[ = \frac{1}{2} \left| -4a^2 - a^2 - \frac{245a^2}{60} \right| \]
\[ = \frac{1}{2} \left| -5a^2 - \frac{245a^2}{60} \right| \]
Finding a common denominator:
\[ = \frac{1}{2} \left| -\frac{300a^2 + 245a^2}{60} \right| = \frac{1}{2} \left| -\frac{545a^2}{60} \right| = \frac{545a^2}{120} \]

### Final Answer
The area of triangle PTQ is:
\[ \text{Area} = \frac{545a^2}{120} \]