JEE Main 2026 (24 January Shift 2)

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Animated Solution for Mathematics - Conic Sections: Let the length of the latus rectum of an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, (a > b)$ , be 30. If its eccentricity is the maximum value of the function $f (t) = - \frac{3}{4} + 2 t - t^{2}$ , then $(a^{2} + b^{2})$ is equal to

Q: Why is the eccentricity $e = 1/4$? I thought eccentricity had to be between 0 and 1, but this seems too small.

Eccentricity $e$ for an ellipse is indeed between 0 and 1. A value of $1/4$ is perfectly valid and indicates an ellipse that is somewhat elongated but not extremely flat.

Q: Can I use the vertex formula $-D/4A$ to find the maximum of the quadratic instead of completing the square?

Absolutely! The vertex formula $-D/4A$ is a fantastic shortcut. For $f(t) = -t^2 + 2t - 3/4$, the vertex occurs at $t = -B/(2A) = -2/(2*-1) = 1$. Plugging $t=1$ into $f(t)$ gives $f(1) = -1 + 2 - 3/4 = 1/4$. Both methods are mathematically sound.

Select Answer:

Visualized Solution

Visualizing the Ellipse and $L R$

Given an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ with $a > b$ .
The length of its Latus Rectum ( $L R$ ) is $30$ .

The Eccentricity Function $f (t)$

The eccentricity $e$ is the maximum value of $f (t) = - \frac{3}{4} + 2 t - t^{2}$ .
This represents a downward-opening parabola.

Maximizing $f (t)$ via Perfect Square

Rewrite $f (t)$ : $f (t) = - (t^{2} - 2 t) - \frac{3}{4}$
Complete the square: $f (t) = - (t^{2} - 2 t + 1 - 1) - \frac{3}{4}$
$f (t) = - (t - 1)^{2} + 1 - \frac{3}{4}$

Finding the Eccentricity $e$

$f (t) = \frac{1}{4} - (t - 1)^{2}$
The maximum value occurs when $(t - 1)^{2} = 0$ , which gives $t = 1$ .
Maximum value = $\frac{1}{4}$ .
Therefore, eccentricity $e = \frac{1}{4}$ .

Applying the Latus Rectum Formula

Formula for Latus Rectum: $L R = \frac{2 b ^{2}}{a}$
Given $L R = 30$ , so $\frac{2 b ^{2}}{a} = 30$
Simplifying: $b^{2} = 15 a$

The Standard Ellipse Identity

Standard relation: $b^{2} = a^{2} (1 - e^{2})$
Substitute $e = \frac{1}{4}$ : $b^{2} = a^{2} (1 - (\frac{1}{4})^{2})$
$b^{2} = a^{2} (1 - \frac{1}{16}) = \frac{15 a ^{2}}{16}$

Equating the Expressions for $b^{2}$

From previous steps: $b^{2} = 15 a$ and $b^{2} = \frac{15 a ^{2}}{16}$
Equating them: $15 a = \frac{15 a ^{2}}{16}$

Solving for $a$ and $a^{2}$

Dividing by $15 a$ (since $a \neq = 0$ ): $1 = \frac{a}{16}$
$a = 16$
Squaring both sides: $a^{2} = 256$

Solving for $b^{2}$

Using the relation $b^{2} = 15 a$
Substitute $a = 16$ : $b^{2} = 15 \times 16$
$b^{2} = 240$

Final Calculation: $a^{2} + b^{2}$

We need to find the value of $a^{2} + b^{2}$ .
$a^{2} + b^{2} = 256 + 240$
$a^{2} + b^{2} = 496$

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The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Solution Diagram

Analyzing the Eccentricity

The eccentricity

e

is defined by the maximum value of the quadratic function

f (t) = - \frac{3}{4} + 2 t - t^{2}

. To find this peak, we complete the square for the expression:

f (t) = - (t^{2} - 2 t) - \frac{3}{4}

By adding and subtracting

1

within the parentheses, we obtain:

f (t) = - (t^{2} - 2 t + 1 - 1) - \frac{3}{4}

f (t) = - (t - 1)^{2} + 1 - \frac{3}{4} = \frac{1}{4} - (t - 1)^{2}

The maximum value of this function occurs when the squared term is zero. Thus, the eccentricity is determined to be

e = \frac{1}{4}

The Geometry of the Latus Rectum

The length of the latus rectum for an ellipse is given by the formula

L R = \frac{2 b ^{2}}{a}

. Given that

L R = 30

, we establish the following relationship:

\frac{2 b ^{2}}{a} = 30 ⟹ b^{2} = 15 a

This equation serves as our primary bridge between the geometric properties and the algebraic parameters of the ellipse.

Solving for the Ellipse Parameters

We utilize the fundamental identity relating the semi-axes and eccentricity:

b^{2} = a^{2} (1 - e^{2})

. Substituting

e = \frac{1}{4}

into this identity yields:

b^{2} = a^{2} (1 - (\frac{1}{4})^{2}) = a^{2} (1 - \frac{1}{16}) = \frac{15 a ^{2}}{16}

Now, we equate our two expressions for

b^{2}

15 a = \frac{15 a ^{2}}{16}

Since

a

represents a physical dimension ($a eq 0$), we divide both sides by

15 a

to find

1 = \frac{a}{16}

, which results in $a = 16$ .

Final Calculation

With the value of

a

determined, we calculate the squares of the semi-axes:

a^{2} = 1 6^{2} = 256

b^{2} = 15 \times 16 = 240

The final requirement is to compute the sum

a^{2} + b^{2}

a^{2} + b^{2} = 256 + 240 = 496

The final result of this geometric derivation is $496$ .

Common Traps & FAQs

Why is the eccentricity

e = 1/4

? I thought eccentricity had to be between 0 and 1, but this seems too small.

Eccentricity

e

for an ellipse is indeed between 0 and 1. A value of

1/4

is perfectly valid and indicates an ellipse that is somewhat elongated but not extremely flat.

Can I use the vertex formula

- D /4 A

to find the maximum of the quadratic instead of completing the square?

Absolutely! The vertex formula

- D /4 A

is a fantastic shortcut. For

f (t) = - t^{2} + 2 t - 3/4

, the vertex occurs at

t = - B / (2 A) = - 2/ (2 * - 1) = 1

. Plugging

t = 1

into

f (t)

gives

f (1) = - 1 + 2 - 3/4 = 1/4

. Both methods are mathematically sound.

Why did we divide by

15 a

in the final step? Is it possible

a = 0

Great catch! In mathematics, we must always be careful when dividing by a variable. However,

a

represents the semi-major axis of an ellipse, which must be a positive length. Therefore,

a

cannot be zero, making our division perfectly valid.

Select Answer:

Visualized Solution

Visualizing the Ellipse and $L R$

The Eccentricity Function $f (t)$

Maximizing $f (t)$ via Perfect Square

Finding the Eccentricity $e$

Applying the Latus Rectum Formula

The Standard Ellipse Identity

Equating the Expressions for $b^{2}$

Solving for $a$ and $a^{2}$

Solving for $b^{2}$

Final Calculation: $a^{2} + b^{2}$

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Eccentricity

The Geometry of the Latus Rectum

Solving for the Ellipse Parameters

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Visualizing the Ellipse and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } LR

The Eccentricity Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(t)

Maximizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(t) via Perfect Square

Finding the Eccentricity .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e

Applying the Latus Rectum Formula

The Standard Ellipse Identity

Equating the Expressions for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2+b2

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Eccentricity

The Geometry of the Latus Rectum

Solving for the Ellipse Parameters

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the eccentricity of an ellipse a2x2​+b2y2​=1,a>b, be 41​. If this ellipse passes through the point (−452​​,3), then a2+b2 is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The length of the latus-rectum of the ellipse, whose foci are (2,5) and (2,−3) and eccentricity is 54​, is

Visualizing the Ellipse and $L R$

The Eccentricity Function $f (t)$

Maximizing $f (t)$ via Perfect Square

Finding the Eccentricity $e$

Equating the Expressions for $b^{2}$

Solving for $a$ and $a^{2}$

Solving for $b^{2}$

Final Calculation: $a^{2} + b^{2}$

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