The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellips

Q: Wait, why is the quadrilateral formed by the tangents a rhombus?

Great observation! Because the ellipse $\frac{x^2}{9} + \frac{y^2}{5} = 1$ is symmetric about both the x-axis and the y-axis, the four tangents at the endpoints of the latus rectum $(\pm ae, \pm \frac{b^2}{a})$ are reflections of each other across these axes. This symmetry ensures that the resulting quadrilateral is a rhombus, which simplifies our area calculation significantly.

Q: Can I just use the formula $\frac{2a^2}{e}$ directly?

Yes, you absolutely can! The area of the quadrilateral formed by the tangents at the ends of the latus rectum is indeed $\frac{2a^2}{e}$. However, in a JEE Advanced setting, it is always safer to derive it or at least understand the geometry behind it, as it helps you verify your answer and avoid silly mistakes.

Q: Why do we only calculate the area of one triangle?

Because of the symmetry we discussed earlier! The four triangles formed by the tangents and the coordinate axes in each quadrant are congruent. By calculating the area of the triangle in the first quadrant and multiplying it by 4, we efficiently find the total area of the rhombus without needing to calculate the other three tangents separately.

Question

The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse $\frac{x^2}{9} + \frac{y^2}{5} = 1$, is

Options:

(A) 27/4 sq. units

(B) 9 sq. units

(C) 27/2 sq. units

(D) 27 sq. units

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the coordinate plane, looking at the beautiful, balanced curve of the ellipse defined by:

$$\frac{x^2}{9} + \frac{y^2}{5} = 1$$

It is not just an equation; it is a path of perfect symmetry. To solve this problem, we must first decode the parameters of this ellipse.

Comparing our equation to the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we immediately identify $a^2 = 9$ and $b^2 = 5$, giving us $a = 3$ and $b = \sqrt{5}$.

### The Latus Rectum: Finding the Anchor Points

Now, we turn our attention to the latus rectum. The latus rectum is the chord passing through the focus, perpendicular to the major axis.

To find its endpoints, we first need the eccentricity $e$. Using the formula $e = \sqrt{1 - \frac{b^2}{a^2}}$, we substitute our values:

$$e = \sqrt{1 - \frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$$

With $e$ in hand, the focus is at $(\pm ae, 0)$, which simplifies to $(\pm 3 	imes \frac{2}{3}, 0) = (\pm 2, 0)$.

The endpoint of the latus rectum in the first quadrant, let us call it $L$, is $(ae, \frac{b^2}{a})$. Substituting our values, we get $L = (2, \frac{5}{3})$. This point is our anchor.

### The Tangent: A Line of Precision

We need the equation of the tangent at $L(2, \frac{5}{3})$. The general equation of a tangent at $(x_1, y_1)$ is:

$$\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$$

Plugging in our coordinates, we get:

$$\frac{x(2)}{9} + \frac{y(\frac{5}{3})}{5} = 1$$

Simplifying this, we get $\frac{2x}{9} + \frac{y}{3} = 1$. This is the line that grazes our ellipse at the latus rectum endpoint.

### The Rhombus: Symmetry in Action

To find the area of the quadrilateral, we need the intercepts of this tangent. Setting $y = 0$, we find the x-intercept:

$$\frac{2x}{9} = 1 \Rightarrow x = \frac{9}{2}$$

Setting $x = 0$, we find the y-intercept:

$$\frac{y}{3} = 1 \Rightarrow y = 3$$

These intercepts define a triangle in the first quadrant with vertices at $(0, 0)$, $(\frac{9}{2}, 0)$, and $(0, 3)$. The area of this triangle is:

$$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \frac{9}{2} 	imes 3 = \frac{27}{4}$$

Because the ellipse is symmetric, the four tangents form a rhombus. The total area is simply 4 times the area of this triangle:

$$4 	imes \frac{27}{4} = \mathbf{27} 	ext{ square units.}$$

### The Takeaway

You have just navigated the geometry of the ellipse. Remember, the formula $\frac{2a^2}{e}$ is a powerful tool, but the true mastery lies in visualizing the symmetry. Keep practicing, and these shapes will become second nature to you.

Animated Solution for Mathematics - Conic Sections: The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1$ , is

Select Answer:

Visualized Solution

Visualizing the Ellipse

Calculating Eccentricity $e$

Focus and Latus Rectum Endpoints

Equation of the Tangent at $L$

Finding the Intercepts

Symmetry and the Quadrilateral

Calculating the Final Area

Summary and Key Takeaway

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Latus Rectum

The Tangent

The Rhombus

The Takeaway

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse 9x2​+5y2​=1, is