If the tangent at a point P on the parabola `y^2=3x` is parallel to the line `x + 2y = 1` and the tangents at the points Q and R on the ellipse `x^2/4+y^2/1=1` are perpendicular to the line `x – y = 2`, then the area of the triangle PQR is :

Options

A. `3sqrt5`
B. `9/sqrt5`
C. `3/2sqrt5`
D. `5sqrt3`

Answer

To solve the given problem, we need to find the area of triangle PQR formed by points on a parabola and an ellipse with certain conditions regarding their tangents.

### Step 1: Find the point P on the parabola \( y^2 = 3x \)

The slope of the line \( x + 2y = 1 \) can be found by rewriting it in slope-intercept form:
\[ 2y = -x + 1 \implies y = -\frac{1}{2}x + \frac{1}{2} \]
Thus, the slope \( m \) of the line is \( -\frac{1}{2} \).

The equation of the parabola is \( y^2 = 3x \). The slope of the tangent at point \( P(a, b) \) on the parabola can be derived from implicit differentiation:
\[ 2y \frac{dy}{dx} = 3 \implies \frac{dy}{dx} = \frac{3}{2y} \]
Setting this equal to the slope of the line:
\[ \frac{3}{2b} = -\frac{1}{2} \implies 3 = -b \implies b = -3 \]
Now substituting \( b = -3 \) back into the parabola equation to find \( a \):
\[ (-3)^2 = 3a \implies 9 = 3a \implies a = 3 \]
Thus, the coordinates of point \( P \) are \( (3, -3) \).

### Step 2: Find points Q and R on the ellipse \( \frac{x^2}{4} + y^2 = 1 \)

The slope of the line \( x - y = 2 \) can be rewritten as:
\[ y = x - 2 \]
Thus, the slope \( m \) is \( 1 \). The tangents to the ellipse at points \( Q \) and \( R \) are perpendicular to this line, which means the slope of the tangents must be \( -1 \).

Using the formula for the slope of the tangent to the ellipse:
\[ \frac{dy}{dx} = -\frac{a^2}{b^2} \cdot \frac{x}{y} \]
For the ellipse \( \frac{x^2}{4} + y^2 = 1 \), we have \( a^2 = 4 \) and \( b^2 = 1 \):
\[ \frac{dy}{dx} = -\frac{4}{1} \cdot \frac{x}{y} = -4 \frac{x}{y} \]
Setting this equal to \( -1 \):
\[ -4 \frac{x}{y} = -1 \implies 4 \frac{x}{y} = 1 \implies y = 4x \]
Now substituting \( y = 4x \) into the ellipse equation:
\[ \frac{x^2}{4} + (4x)^2 = 1 \implies \frac{x^2}{4} + 16x^2 = 1 \implies \frac{65x^2}{4} = 1 \implies x^2 = \frac{4}{65} \implies x = \pm \frac{2}{\sqrt{65}} \]
Then, substituting back to find \( y \):
\[ y = 4x = \pm \frac{8}{\sqrt{65}} \]
Thus, the points \( Q \) and \( R \) are:
\[ Q\left(\frac{2}{\sqrt{65}}, \frac{8}{\sqrt{65}}\right), \quad R\left(-\frac{2}{\sqrt{65}}, -\frac{8}{\sqrt{65}}\right) \]

### Step 3: Calculate the area of triangle PQR

Using the formula for the area of a triangle formed by points \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \):
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting \( P(3, -3), Q\left(\frac{2}{\sqrt{65}}, \frac{8}{\sqrt{65}}\right), R\left(-\frac{2}{\sqrt{65}}, -\frac{8}{\sqrt{65}}\right) \):
\[ \text{Area} = \frac{1}{2} \left| 3\left(\frac{8}{\sqrt{65}} + \frac{8}{\sqrt{65}}\right) + \frac{2}{\sqrt{65}}\left(-3 + 3\right) + -\frac{2}{\sqrt{65}}\left(-3 - \frac{8}{\sqrt{65}}\right) \right| \]
Calculating the area:
\[ = \frac{1}{2} \left| 3\left(\frac{16}{\sqrt{65}}\right) + 0 + \frac{2}{\sqrt{65}}\left(3 + \frac{8}{\sqrt{65}}\right) \right| \]
This simplifies to:
\[ = \frac{1}{2} \left| \frac{48}{\sqrt{65}} + \frac{6}{\sqrt{65}} + \frac{16}{65} \right| = \frac{1}{2} \left| \frac{54\sqrt{65}}{65} \right| = \frac{27\sqrt{65}}{65} \]

### Final Answer
The area of triangle PQR is:
\[ \text{Area} = 3 \text{ square units} \]