If the minimum area of the triangle formed by a tangent to the ellipse x²/b² + y²/4a² = 1 and the co

Q: Wait, why did we rewrite the ellipse equation as $\frac{x^2}{b^2} + \frac{y^2}{(2a)^2} = 1$? Can't we just use the standard form?

That is a fantastic observation! You absolutely can use the standard form, but rewriting it as $\frac{x^2}{b^2} + \frac{y^2}{(2a)^2} = 1$ is a strategic move. It explicitly reveals the semi-axes lengths as $b$ and $2a$. This makes the parametric substitution $(b \cos \theta, 2a \sin \theta)$ intuitive and prevents silly algebraic errors when you plug these values into the tangent equation. In JEE, clarity is speed!

Q: Why do we use the eccentric angle $\theta$ instead of just solving for $y$ in terms of $x$?

Great question! If you solve for $y$ in terms of $x$, you will encounter square roots and messy derivatives. By using the parametric coordinate $\theta$, we transform a calculus problem into a trigonometric one. Trigonometric identities are much more elegant and easier to manipulate for finding minima than dealing with radical functions. Always choose the path that simplifies the algebra!

Q: How do we know the minimum area occurs when $\sin 2\theta = 1$?

Think about the structure of our area expression: $\text{Area} = \frac{2ab}{\sin 2\theta}$. Since $2ab$ is a constant, the only way to minimize the entire fraction is to maximize the denominator. The sine function, $\sin 2\theta$, has a maximum value of $1$. Therefore, when $\sin 2\theta = 1$, the denominator is at its largest, making the overall area as small as possible. It is all about controlling the variables!

Question

If the minimum area of the triangle formed by a tangent to the ellipse $\frac{x^2}{b^2} + \frac{y^2}{4a^2} = 1$ and the co-ordinate axis is $kab$, then $k$ is equal to .

Sigma Percentile · Accepted Answer

### The Geometry of the Ellipse: A Journey into Optimization

Welcome, future engineer! Today, we are going to peel back the layers of a classic coordinate geometry problem. We are looking at an ellipse, but not just any ellipse—one that demands we understand the relationship between its tangent lines and the coordinate axes.

This problem is a beautiful example of how we can transform a seemingly complex geometric setup into a clean, elegant trigonometric optimization. Let us dive in.

### Phase 1: Visualizing the Ellipse

We begin with the given equation:

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

The first thing a seasoned problem solver does is look for the structure. We see $4a^2$ in the denominator of the $y$-term.

Let us rewrite this as:

$$
\frac{x^2}{b^2} + \frac{y^2}{(2a)^2} = 1
$$

Now, the geometry is laid bare. We have an ellipse with semi-axes $b$ and $2a$. Visualizing this on the Cartesian plane, we see a shape stretched along the $y$-axis. This clarity is our foundation.

### Phase 2: The Parametric Dance

Now, we need to draw a tangent. If we try to work with $y$ as a function of $x$, we will be drowning in square roots and derivatives.

Instead, let us use the power of parametric coordinates. Any point $P$ on this ellipse can be elegantly described by the eccentric angle $	heta$ as:

$$
P(b \cos 	heta, 2a \sin 	heta)
$$

This choice is not just convenient; it is strategic. It allows us to handle the tangent equation with grace.

### Phase 3: The Tangent's Geometry

With our point $P$ defined, we recall the standard equation of a tangent to an ellipse at $(x_1, y_1)$, which is:

$$
\frac{x x_1}{b^2} + \frac{y y_1}{(2a)^2} = 1
$$

Substituting our parametric coordinates, we get:

$$
\frac{x(b \cos 	heta)}{b^2} + \frac{y(2a \sin 	heta)}{(2a)^2} = 1
$$

Watch closely as the terms simplify. The $b$ cancels, the $2a$ cancels, and we are left with a beautiful, linear equation:

$$
\frac{x \cos 	heta}{b} + \frac{y \sin 	heta}{2a} = 1
$$

This is the line that cuts across our axes.

### Phase 4: The Intercepts and the Area

To find the area of the triangle formed by this tangent and the axes, we need the intercepts. Setting $y=0$ gives us the $x$-intercept $A = \frac{b}{\cos 	heta}$.

Setting $x=0$ gives us the $y$-intercept $B = \frac{2a}{\sin 	heta}$. We have created a right-angled triangle with vertices at the origin, $A$, and $B$.

The area is simply $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$, which becomes:

$$
	ext{Area} = \frac{1}{2} 	imes \left(\frac{b}{\cos 	heta}ight) 	imes \left(\frac{2a}{\sin 	heta}ight)
$$

### Phase 5: The Climax of Optimization

Simplifying this, we get:

$$
	ext{Area} = \frac{ab}{\sin 	heta \cos 	heta}
$$

Here is the moment of insight: we recognize the denominator as half of the double-angle identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$. Thus, our area becomes:

$$
	ext{Area} = \frac{2ab}{\sin 2	heta}
$$

To minimize this area, we must maximize the denominator. Since the maximum value of $\sin 2	heta$ is $1$, the minimum area is simply **$2ab$**.

Comparing this to the given $kab$, we find that **$k=2$**. See how the complexity melted away? That is the beauty of mathematics—when you choose the right tools, the path to the solution becomes a clear, logical journey.

Animated Solution for Mathematics - Conic Sections: If the minimum area of the triangle formed by a tangent to the ellipse $\frac{x ^{2}}{b ^{2}} + \frac{y ^{2}}{4 a ^{2}} = 1$ and the co-ordinate axis is $k ab$ , then $k$ is equal to .

Enter Numerical Value:

Visualized Solution

Visualizing the Ellipse

Defining a Parametric Point

Equation of the Tangent

Simplifying the Tangent Equation

Finding the x-intercept

Finding the y-intercept

Area of Triangle $O A B$

Simplifying the Area Expression

Condition for Minimum Area

Calculating $k$

The Sigma Insight: Equation of Tangent and Normal

The Geometry of the Ellipse

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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