Consider the ellipse x²/9 + y²/4 = 1. Let S(p, q) be a point in the first quadrant such that p²/9 +

Q: Wait, why do we assume the tangent at the minor axis is $y=2$?

Great question! Look at the ellipse equation: $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$. The semi-minor axis $b$ is 2. The endpoints of the minor axis are $(0, 2)$ and $(0, -2)$. Since the problem states $S$ is in the first quadrant, the tangent must be the horizontal line passing through the upper endpoint, which is $(0, 2)$. Thus, the tangent is simply $y=2$.

Q: Why is the height of $\triangle ORT$ just $|2\sin\theta|$?

Think about the geometry. The vertex $R$ is $(3, 0)$ and $O$ is $(0, 0)$. The base $OR$ lies entirely on the x-axis. The height of any triangle is the perpendicular distance from the third vertex ($T$) to the line containing the base. Since the base is on the x-axis, the height is simply the absolute value of the y-coordinate of $T$. Given $T(3\cos\theta, 2\sin\theta)$, the y-coordinate is $2\sin\theta$.

Q: Can I use the slope form $y = mx + c$ instead of parametric coordinates?

You absolutely can! You could use the condition of tangency $c^2 = a^2m^2 + b^2$. However, parametric coordinates are often more elegant for ellipse problems involving points of contact. They reduce the number of variables and make the area calculation much cleaner. Stick with the parametric approach; it's a powerful tool in your JEE arsenal.

Question

Consider the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$. Let $S(p, q)$ 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 centre of the ellipse. If the area of the triangle $	riangle ORT$ is $\frac{3}{2}$, then which of the following options is correct ?

Options:

(A) $q = 2, p = 3\sqrt{3}$

(B) $q = 2, p = 4\sqrt{3}$

(C) $q = 1, p = 5\sqrt{3}$

(D) $q = 1, p = 6\sqrt{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the JEE path. Today, we are not just solving a problem; we are peeling back the layers of an ellipse to reveal the elegant geometry hidden beneath.

When you look at the equation

$$\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$$

do not just see numbers. See a shape with a center at the origin $O(0, 0)$, a semi-major axis of $3$, and a semi-minor axis of $2$. This is our canvas.

### Phase 1: The First Tangent - The Hidden Anchor

The problem introduces a point $S(p, q)$ in the first quadrant. We are told that a tangent from $S$ meets the ellipse at the minor axis.

The minor axis endpoints are $(0, 2)$ and $(0, -2)$. Since $S$ is in the first quadrant, the tangent must be the horizontal line passing through $(0, 2)$.

Therefore, the tangent is $y = 2$. Because $S$ lies on this tangent, we have immediately unlocked the first coordinate of our point: $q = 2$. Our point $S$ is now $(p, 2)$.

### Phase 2: The Geometry of $T$ and the Area Constraint

Now, the plot thickens. A second tangent is drawn from $S$ to a point $T$ in the fourth quadrant. To handle this, we invoke the power of parametric coordinates.

Any point on our ellipse can be represented as $(3\cos	heta, 2\sin	heta)$. Let $T$ be $(3\cos	heta, 2\sin	heta)$.

We are given that the area of $	riangle ORT$ is $\frac{3}{2}$. The base $OR$ lies on the x-axis, stretching from the origin $(0, 0)$ to the vertex $R(3, 0)$, giving a length of $3$.

The height of the triangle is the perpendicular distance from $T$ to the x-axis, which is $|2\sin	heta|$. Using the area formula $	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$, we get:

$$	ext{Area} = \frac{1}{2} 	imes 3 	imes |2\sin	heta| = 3|\sin	heta|$$

Equating this to the given area of $\frac{3}{2}$, we find $3|\sin	heta| = \frac{3}{2}$, which simplifies to $|\sin	heta| = \frac{1}{2}$. Since $T$ is in the fourth quadrant, $\sin	heta = -\frac{1}{2}$.

Using the identity $\cos^2	heta + \sin^2	heta = 1$, we find $\cos	heta = \frac{\sqrt{3}}{2}$.

### Phase 3: The Final Convergence

Now we know the exact coordinates of $T$. Substituting our values:

$$T = \left(3 	imes \frac{\sqrt{3}}{2}, 2 	imes \left(-\frac{1}{2}ight)ight) = \left(\frac{3\sqrt{3}}{2}, -1ight)$$

The equation of the tangent at any point $(x_1, y_1)$ on the ellipse is $\frac{xx_1}{9} + \frac{yy_1}{4} = 1$. Substituting our point $T$, we get:

$$\frac{x \cdot \left(\frac{3\sqrt{3}}{2}ight)}{9} + \frac{y \cdot (-1)}{4} = 1$$

Simplifying this, we obtain $\frac{x\sqrt{3}}{6} - \frac{y}{4} = 1$. Since $S(p, 2)$ lies on this line, we substitute $x = p$ and $y = 2$:

$$\frac{p\sqrt{3}}{6} - \frac{2}{4} = 1 \Rightarrow \frac{p\sqrt{3}}{6} - \frac{1}{2} = 1$$

$$\frac{p\sqrt{3}}{6} = \frac{3}{2}$$

Multiplying by $6$, we get $p\sqrt{3} = 9$. Finally, **$p = 3\sqrt{3}$**.

We have arrived at our destination: **$q = 2$** and **$p = 3\sqrt{3}$**. By breaking the problem into geometric phases, we turned a complex problem into a logical sequence of steps.

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Visualized Solution

The Ellipse and its Vertices

The First Tangent

Finding the y-coordinate of $S$

The Second Tangent and Point $T$

Area of Triangle $O R T$

Equating the Area

Solving for $θ$

Exact Coordinates of $T$

Equation of Tangent at $T$

Substituting $S$ to find $p$

Final Calculation for $p$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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Visualized Solution

The Ellipse and its Vertices

The First Tangent

Finding the y-coordinate of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

The Second Tangent and Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

Area of Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ORT

Equating the Area

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

Exact Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

Equation of Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Final Calculation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p