Consider the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$. Let S(p,q) q) be a point in the first quadrant such that $\frac{p^2}{9}+\frac{q^2}{4}>1$. Two tangents are drawn from S to the ellipse, of which one meets the ellipse at one end point of the minor axis and the other meets the ellipse at a point T in the fourth quadrant. Let R be the vertex of the ellipse with positive x-coordinate and O be the center of the ellipse. If the area of the triangle ∆ORT is 3/2, then which of the following options is correct?

The correct option is q = 2, p = 3√3.

Here is the step-by-step derivation:

Step 1: Analyze the Ellipse and Coordinates
The given equation of the ellipse is:
$\frac{x^2}{9}+\frac{y^2}{4}=1$
Comparing this with the standard equation of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we find:
$a^2=9\Rightarrow a=3$
$b^2=4\Rightarrow b=2$
The center of the ellipse is $O(0,0)$. The vertex with a positive x-coordinate is $R(3,0)$.

Step 2: Determine the Coordinates of Point S(p, q)
Let $S(p,q)$ be a point in the first quadrant, so $p>0$ and $q>0$. We are given that one of the tangents from S to the ellipse meets the ellipse at an end point of the minor axis.
The end points of the minor axis are $(0,2)$ and $(0,-2)$.
The tangent to the ellipse at $(0,2)$ is the horizontal line:
$y=2$
Since S lies in the first quadrant $(q>0)$, it must lie on this line $y=2$, which gives:
$q=2$
Thus, the point S is $S(p,2)$.

Step 3: Analyze the Second Tangent and Point of Contact T
Let the point of contact of the second tangent from S to the ellipse be $T(x_1,y_1)$ in the fourth quadrant, which implies $x_1>0$ and $y_1<0$.
The equation of the tangent at $T(x_1,y_1)$ is:
$\frac{x{x}_1}{9}+\frac{y{y}_1}{4}=1$
Since this tangent passes through $S(p,2)$, we can substitute these coordinates:
$\frac{p{x}_1}{9}+\frac{2y_1}{4}=1\Rightarrow \frac{p{x}_1}{9}+\frac{y_1}{2}=1$ (Equation 1)

Step 4: Use the Area of Triangle ORT to Find $y_1$
The vertices of triangle ORT are $O(0,0)$, $R(3,0)$, and $T(x_1,y_1)$.
The area of triangle ORT is given by:
$\text{Area}=\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times OR\times |y_1|$
Since T lies in the fourth quadrant, $y_1<0$, meaning $|y_1|=-y_1$. Thus:
$\frac{3}{2}=\frac{1}{2}\times 3\times (-y_1)\Rightarrow y_1=-1$

Step 5: Find $x_1$
Since $T(x_1,y_1)$ lies on the ellipse, we have:
$\frac{x_1^2}{9}+\frac{y_1^2}{4}=1$
Substitute $y_1=-1$ into this relation:
$\frac{x_1^2}{9}+\frac{1}{4}=1\Rightarrow \frac{x_1^2}{9}=\frac{3}{4}\Rightarrow x_1^2=\frac{27}{4}$
Since $x_1>0$ in the fourth quadrant, we get:
$x_1=\frac{3\sqrt{3}}{2}$

Step 6: Calculate p
Substitute the values of $x_1$ and $y_1$ into Equation 1:
$\frac{p·\frac{3\sqrt{3}}{2}}{9}+\frac{-1}{2}=1$
Simplify the equation:
$\frac{p\sqrt{3}}{6}-\frac{1}{2}=1\Rightarrow \frac{p\sqrt{3}}{6}=\frac{3}{2}$
Multiply both sides by 6:
$p\sqrt{3}=9\Rightarrow p=\frac{9}{\sqrt{3}}=3\sqrt{3}$
Therefore, the values are q = 2 and p = 3√3.