Let T be the tangent to the ellipse E : x² + 4y² = 5 at the point P(1, 1). If the area of the region

Q: Wait, why is the tangent the upper curve and the ellipse the lower curve in this region?

Great observation! If you sketch the ellipse and the tangent at $(1, 1)$, you will see that the tangent line lies above the ellipse for $x > 1$. Since we are integrating from $x = 1$ to $x = \sqrt{5}$, the tangent is indeed the upper boundary. Always sketch the curves if you are unsure about the sign of the area!

Q: Why do we use $\cos^{-1}$ in the final answer when the integral formula gives $\sin^{-1}$?

That is a classic JEE twist! We use the trigonometric identity $\sin^{-1}(\theta) + \cos^{-1}(\theta) = \frac{\pi}{2}$. When we evaluate the integral, we get a $\sin^{-1}$ term. By converting it to $\cos^{-1}$, we match the specific format requested in the question, which is a common way to test your comfort with inverse trigonometric identities.

Q: Is it possible to solve this using geometry instead of integration?

Yes, you could! The area is essentially the area of a right-angled triangle formed by the tangent and the axes, minus the area of the sector of the ellipse. However, for JEE Advanced, mastering the integration approach is crucial because it is more robust and less prone to geometric errors. Stick with the integral; it is the most reliable path.

Question

Let $T$ be the tangent to the ellipse $E : x^2 + 4y^2 = 5$ at the point $P(1, 1)$. If the area of the region bounded by the tangent $T$, ellipse $E$, lines $x = 1$ and $x = \sqrt{5}$ is $\alpha\sqrt{5} + \beta + \gamma \cos^{-1} \left( \frac{1}{\sqrt{5}} \right)$, then $|\alpha + \beta + \gamma|$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at the elegant curve of the ellipse $E : x^2 + 4y^2 = 5$. It is a beautiful, symmetric shape, but today, we are not just admiring its form; we are dissecting it.

We are focusing on a specific point $P(1, 1)$ on this curve. First, let us verify that $P$ truly belongs to our ellipse. Substituting $x=1$ and $y=1$ into the equation $x^2 + 4y^2 = 5$, we get $1^2 + 4(1)^2 = 5$, which is $1 + 4 = 5$. It fits perfectly.

Now, imagine drawing a tangent line $T$ at this point $P$. This line is the local linear approximation of our curve, and it is the key to the region we are about to measure.

### Defining the Tangent

To find the equation of the tangent $T$ at $(x_1, y_1) = (1, 1)$, we use the standard tangent formula for an ellipse: $xx_1 + 4yy_1 = 5$. Substituting our point $(1, 1)$, we get $x(1) + 4y(1) = 5$, which simplifies to $x + 4y = 5$.

Rearranging this, we find the explicit equation for our tangent line:
$$y = \frac{5-x}{4}$$
This linear equation will serve as our upper boundary for the integration.

### The Bounded Region

Now, look at the region bounded by the tangent $T$, the ellipse $E$, and the vertical lines $x = 1$ and $x = \sqrt{5}$. The tangent line acts as the ceiling, and the ellipse acts as the floor.

The area $A$ is the integral of the difference between these two functions:
$$A = \int_{1}^{\sqrt{5}} \left( \frac{5-x}{4} - \frac{\sqrt{5-x^2}}{2} ight) dx$$
Do not let this integral intimidate you. We can break it into two manageable pieces: the linear part and the radical part.

### The Linear and Radical Integration

First, the linear part: $\int_{1}^{\sqrt{5}} \frac{5-x}{4} dx$. This is a simple polynomial integration. It becomes:
$$\left[ \frac{5x}{4} - \frac{x^2}{8} ight]_{1}^{\sqrt{5}}$$
Evaluating this at the limits, we get:
$$\left( \frac{5\sqrt{5}}{4} - \frac{5}{8} ight) - \left( \frac{5}{4} - \frac{1}{8} ight) = \frac{5\sqrt{5}}{4} - \frac{7}{4}$$

Next, the radical part: $\int_{1}^{\sqrt{5}} \frac{\sqrt{5-x^2}}{2} dx$. This is where we use the standard integral formula $\int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$.

Applying this, we get:
$$\frac{1}{2} \left[ \frac{x}{2}\sqrt{5-x^2} + \frac{5}{2}\sin^{-1}\left(\frac{x}{\sqrt{5}}ight) ight]_{1}^{\sqrt{5}}$$
Evaluating at the limits, we find the area of the radical part to be:
$$\frac{5\pi}{8} - \frac{1}{2} - \frac{5}{4}\sin^{-1}\left(\frac{1}{\sqrt{5}}ight)$$

### The Final Synthesis

Finally, we combine these results:
$$A = \left(\frac{5\sqrt{5}}{4} - \frac{7}{4}ight) - \left(\frac{5\pi}{8} - \frac{1}{2} - \frac{5}{4}\sin^{-1}\left(\frac{1}{\sqrt{5}}ight)ight)$$
With a bit of algebraic grace, we use the identity $\frac{\pi}{2} - \sin^{-1}(	heta) = \cos^{-1}(	heta)$ to transform our expression into:
$$A = \frac{5}{4}\sqrt{5} - \frac{5}{4} - \frac{5}{4}\cos^{-1}\left(\frac{1}{\sqrt{5}}ight)$$

Comparing this to the form $\alpha\sqrt{5} + \beta + \gamma \cos^{-1} \left( \frac{1}{\sqrt{5}} ight)$, we identify $\alpha = \frac{5}{4}$, $\beta = -\frac{5}{4}$, and $\gamma = -\frac{5}{4}$.

The absolute sum is:
$$|\alpha + \beta + \gamma| = \left|\frac{5}{4} - \frac{5}{4} - \frac{5}{4}ight| = \mathbf{1.25}$$
You have successfully navigated the geometry and the calculus to find the answer. Keep this momentum going!

Enter Numerical Value:

Visualized Solution

Visualize the Ellipse and Point $P$

Equation of Tangent $T$

Identify the Bounded Region

Set up the Definite Integral

Integrate the Linear Part

Integrate the Radical Part

Evaluate Limits for Radical Part

Combine and Simplify

Identify $α, β, γ$

Final Calculation

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Defining the Tangent

The Bounded Region

The Linear and Radical Integration

The Final Synthesis

Similar Questions

The area of the region, inside the ellipse $x^{2} + 4 y^{2} = 4$ and outside the region bounded by the curves $y = ∣ x ∣ - 1$ and $y = 1 - ∣ x ∣$ , is :

The area of the region ${(x, y) : ∣ x - 1∣ \leq y \leq 5 - x^{2}}$ is equal to:

The area (in square units) of the region enclosed by the ellipse $x^{2} + 3 y^{2} = 18$ in the first quadrant below the line $y = x$ is

The area of the region described by $A = {(x, y) : x^{2} + y^{2} \leq 1 and y^{2} \leq 1 - x}$ is:

The area (in sq. units) in the first quadrant bounded by the parabola, $y = x^{2} + 1$ , the tangent to it at the point $(2, 5)$ and the coordinate axes is:

Find the area bounded by the curves, $x^{2} + y^{2} = 25, 4 y = 4 - x^{2}$ and $x = 0$ above the $x$ -axis.

The parabola $y^{2} = 4 x$ divides the area of the circle $x^{2} + y^{2} = 5$ in two parts. The area of the smaller part is equal to:

Let $Δ$ be the area of the region ${(x, y) \in R^{2} : x^{2} + y^{2} \leq 21, y^{2} \leq 4 x, x \geq 1}$ . Then $\frac{1}{2} (Δ - 21 sin^{- 1} \frac{2}{7})$ is equal to

The area of the region bounded by the curves $x (1 + y^{2}) = 1$ and $y^{2} = 2 x$ is: