Let an ellipse with centre (1, 0) and latus rectum of length 1/2 have its major axis along x-axis. I

Q: Wait, why is the angle at the focus 30 degrees and not 60 degrees?

Great question! The problem states the *entire* minor axis subtends 60 degrees at the focus. Since the ellipse is symmetric about the major axis, the line from the focus to the center bisects this angle. Thus, the triangle formed by the center, the focus, and the top endpoint of the minor axis has an angle of 60/2 = 30 degrees at the focus.

Q: Why can't we just use the focus-directrix property here?

You absolutely could! The focus-directrix property is a powerful tool. However, in this specific problem, the geometric approach using the right-angled triangle formed by the center, focus, and minor axis endpoint is much more direct and less prone to algebraic errors. Always look for the path of least resistance in geometry!

Q: Why did we reject the solution a = 0?

In mathematics, we often get multiple roots for an equation. However, in physics and geometry, we must apply physical constraints. 'a' represents the length of the semi-major axis of an ellipse. A length must be a positive, non-zero value for the shape to exist. Therefore, a = 0 is physically impossible.

Question

Let an ellipse with centre $(1, 0)$ and latus rectum of length $\frac{1}{2}$ have its major axis along x-axis. If its minor axis subtends an angle $60^\circ$ at the foci, then the square of the sum of the lengths of its minor and major axes is equal to _______.

Sigma Percentile · Accepted Answer

### The Geometry of Elegance: Unlocking the Ellipse

My dear student, welcome to a beautiful exploration of conic sections. Today, we are not just solving for a number; we are peeling back the layers of an ellipse to understand its geometric soul.

Imagine you are standing on the Cartesian plane. We have an ellipse centered at $(1, 0)$, stretching its major axis along the $x$-axis. It feels like a simple setup, but within this simplicity lies a classic JEE challenge that tests your ability to bridge the gap between pure geometry and algebraic rigor.

### Phase 1: The Foundation

Let us define our variables. We have the semi-major axis $a$ and the semi-minor axis $b$. The center is at $(1, 0)$, but remember, the shape of an ellipse is defined by its dimensions, not its position.

The problem gives us a crucial clue: the length of the latus rectum is $\frac{1}{2}$. Do you recall the standard formula? It is $\frac{2b^2}{a}$.

By setting the equation:

$$ \frac{2b^2}{a} = \frac{1}{2} $$

We immediately find a vital relationship: $4b^2 = a$. Let us hold onto this as our first anchor, equation (1).

### Phase 2: The Geometric Insight

Now, let us visualize the focus. The problem tells us the minor axis subtends an angle of $60^\circ$ at the focus.

If we draw a triangle connecting the center $O'(1, 0)$, the focus $S$, and the upper endpoint of the minor axis $B(1, b)$, we create a right-angled triangle $	riangle SBO'$. Because of the symmetry of the ellipse, the line from the focus to the center bisects the angle subtended by the minor axis.

Thus, $\angle BSO' = 30^\circ$. This is the key that unlocks the door!

In this right-angled triangle, the base $O'S$ is the distance from the center to the focus, which is $ae$. The height $O'B$ is simply $b$. Using trigonometry, we have:

$$ 	an 30^\circ = \frac{b}{ae} $$

Since $	an 30^\circ = \frac{1}{\sqrt{3}}$, we get $\frac{1}{\sqrt{3}} = \frac{b}{ae}$, which simplifies to $ae = \sqrt{3}b$.

### Phase 3: The Grand Synthesis

We are now in the home stretch. We have $ae = \sqrt{3}b$. Squaring both sides gives us $a^2e^2 = 3b^2$.

Now, we invoke the fundamental identity of an ellipse: $b^2 = a^2(1 - e^2)$. Expanding this, we get $b^2 = a^2 - a^2e^2$, or $a^2e^2 = a^2 - b^2$.

By substituting our squared relation $a^2e^2 = 3b^2$ into this identity, we get:

$$ 3b^2 = a^2 - b^2 $$

This simplifies beautifully to $a^2 = 4b^2$. This is our second anchor, equation (2).

### Phase 4: The Final Victory

Look at what we have achieved. From equation (1), we have $a = 4b^2$. From equation (2), we have $a^2 = 4b^2$.

Substituting the first into the second, we get $a^2 = a$. This leads to $a(a - 1) = 0$.

Since $a$ must be a positive length, $a = 1$. Consequently, $4b^2 = 1$, so $b^2 = \frac{1}{4}$, meaning $b = \frac{1}{2}$.

The major axis length is $2a = 2$, and the minor axis length is $2b = 1$. The sum is $2 + 1 = 3$.

The problem asks for the square of this sum, so $3^2 = 9$. We have arrived at our destination. The final answer is **9**.

Enter Numerical Value:

Visualized Solution

Visualizing the Ellipse

Latus Rectum Condition

Relating $a$ and $b^{2}$

Angle at the Focus

Trigonometry in $△ S B O^{'}$

Expressing $a e$ in terms of $b$

Squaring the Relation

The Eccentricity Identity

Relating $a^{2}$ and $b^{2}$

Solving for $a$

Finding the value of $a$

Finding the value of $b$

Lengths of the Axes

Final Calculation

The Sigma Insight: Foci, Directrices, and Eccentricity

The Geometry of Elegance

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

If distance between the foci of an ellipse is 6 and distance between its directrices is 12, then length of its latus rectum is

Let $f (x) = x^{2} + 9, g (x) = \frac{x}{x - 9}$ and $a = f \circ g (10), b = g \circ f (3)$ . If $e$ and $l$ denote the eccentricity and the length of the latus rectum of the ellipse $\frac{x ^{2}}{a} + \frac{y ^{2}}{b} = 1$ , then $8 e^{2} + l^{2}$ is equal to.

If the length of the minor axis of ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is :

If the length of the minor axis of an ellipse is equal to one fourth of the distance between the foci, then the eccentricity of the ellipse is :

Let $H : \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1, a > 0, b > 0$ , be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is $4 (22 + 14)$ . If the eccentricity $H$ is $\frac{11}{2}$ , then value of $a^{2} + b^{2}$ is equal to ______.

Let the product of the focal distances of the point $P (4, 23)$ on the hyperbola $H : \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ be 32. Let the length of the conjugate axis of $H$ be $p$ and the length of its latus rectum be $q$ . Then $p^{2} + q^{2}$ is equal to _______

Let $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1 (a > b)$ be a given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function, $ϕ (t) = \frac{5}{12} + t - t^{2}$ , then $a^{2} + b^{2}$ is equal to :

Let H be the hyperbola, whose foci are $(1 \pm 2, 0)$ and eccentricity is $2$ . Then the length of its latus rectum is

An ellipse has $O B$ as semi minor axis, $F$ and $F^{'}$ its focii and the angle $F B F^{'}$ is a right angle. Then the eccentricity of the ellipse is