Let x²/a² + y²/b² = 1(a b) be a given ellipse, length of whose latus rectum is 10. If its eccentri

Q: Wait, why do we assume the ellipse is horizontal?

The problem explicitly states $a > b$. In the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, if $a > b$, the major axis lies along the x-axis. If $b > a$, it would be vertical. Always check this condition first!

Q: Do I really need calculus to find the maximum of $\phi(t)$?

Not at all! While you *could* use derivatives, the vertex formula $t = -\frac{B}{2A}$ is much faster for quadratics. It's a classic JEE shortcut that saves precious time.

Q: What if I got $e > 1$?

That's a great red flag! For an ellipse, the eccentricity $e$ must always be between $0$ and $1$. If you ever calculate $e > 1$, pause and re-check your vertex calculation or your function evaluation.

Question

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, $\phi(t) = \frac{5}{12} + t - t^2$, then $a^2 + b^2$ is equal to :

Options:

(A) 135

(B) 116

(C) 126

(D) 145

Sigma Percentile · Accepted Answer

### The Geometry of the Ellipse

Imagine you are standing before the elegant, balanced curve of an ellipse. It is not just a shape; it is a path defined by the sum of distances to two fixed points, the foci.

We are given the standard equation:

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

The problem gives us a vital clue: $a > b$. This tells us immediately that our ellipse is stretched horizontally along the $x$-axis. It is a wide, graceful shape, not a tall, narrow one. This geometric orientation is our anchor.

### The Latus Rectum: A Geometric Anchor

The problem mentions the latus rectum, a chord passing through the focus perpendicular to the major axis. For our horizontal ellipse, the length of this chord is a standard result:

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

We are told this length is exactly $10$. By setting $\frac{2b^2}{a} = 10$, we find a beautiful, simple relationship:

$$b^2 = 5a$$

This is a powerful tool we will use to simplify our final equation later. Keep this relation safe in your toolkit.

### The Parabolic Peak: Finding Eccentricity

Next, we encounter a function $\phi(t) = \frac{5}{12} + t - t^2$. The problem states that the eccentricity $e$ of our ellipse is the maximum value of this function.

Look at the structure of $\phi(t)$. It is a quadratic, specifically a downward-opening parabola because the coefficient of $t^2$ is negative. Such a function reaches its peak at its vertex.

For any quadratic $At^2 + Bt + C$, the vertex occurs at $t = -\frac{B}{2A}$. Here, $A = -1$ and $B = 1$. Thus, the maximum occurs at:

$$t = -\frac{1}{2(-1)} = \frac{1}{2}$$

Now, we calculate the maximum value by substituting $t = \frac{1}{2}$ back into $\phi(t)$:

$$\phi\left(\frac{1}{2}ight) = -\left(\frac{1}{2}ight)^2 + \frac{1}{2} + \frac{5}{12} = -\frac{1}{4} + \frac{1}{2} + \frac{5}{12}$$

Finding a common denominator of $12$, we get:

$$-\frac{3}{12} + \frac{6}{12} + \frac{5}{12} = \frac{8}{12} = \frac{2}{3}$$

Thus, our eccentricity $e$ is $\frac{2}{3}$.

### The Grand Unification

We have reached the synthesis phase. We know $b^2 = 5a$ and $e = \frac{2}{3}$. We also know the fundamental identity for an ellipse: $b^2 = a^2(1 - e^2)$.

Let us substitute our known values into this identity:

$$5a = a^2\left(1 - \left(\frac{2}{3}ight)^2ight)$$

$$5a = a^2\left(1 - \frac{4}{9}ight)$$

$$5a = a^2\left(\frac{5}{9}ight)$$

Since $a$ is a length, $a 
eq 0$, so we can safely divide both sides by $5a$:

$$1 = \frac{a}{9} \Rightarrow a = 9$$

With $a = 9$, we find $a^2 = 81$. Now, return to our relation $b^2 = 5a$:

$$b^2 = 5 	imes 9 = 45$$

The problem asks for the sum $a^2 + b^2$. Adding our results, we get $81 + 45 = 126$.

We have navigated the geometry, analyzed the function, and unified the results to find our answer. The final result is **126**.

Select Answer:

Visualized Solution

Visualizing the Ellipse

The Latus Rectum Formula

Setting up the $L R$ Equation

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

Introducing the Function $ϕ (t)$

Finding the Maxima Point

Calculating the Maximum Value

Determining Eccentricity $e$

The Eccentricity Relation

Substitution of Knowns

Simplifying the Equation

Solving for $a$

Finding $b^{2}$

The Final Calculation

The Sigma Insight: Foci, Directrices, and Eccentricity

The Geometry of the Ellipse

The Latus Rectum

The Parabolic Peak

The Grand Unification

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