JEE Main 2025 April

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Animated Solution for Mathematics - Conic Sections: Let the length of a latus rectum of an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ be 10 . If its eccentricity is the minimum value of the function $f (t) = t^{2} + t + \frac{11}{12}, t \in R$ , then $a^{2} + b^{2}$ is equal to :

Q: Wait, why do we use the vertex formula $t = -\frac{b}{2a}$? Can't we just guess the value?

Guessing is the enemy of precision! The function $f(t) = t^2 + t + \frac{11}{12}$ is a parabola opening upwards. Its minimum is strictly defined at the vertex. Using the formula $t = -\frac{b}{2a}$ is the mathematically rigorous way to ensure you find the absolute minimum, not just an approximation.

Q: Why is the latus rectum formula $\frac{2b^2}{a}$ and not $\frac{2a^2}{b}$?

Great question! In the standard ellipse equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we assume $a > b$. The latus rectum is the chord passing through the focus perpendicular to the major axis. The formula $\frac{2b^2}{a}$ is specific to the major axis being along the x-axis. Always check your orientation!

Q: Why do we reject the possibility of $a=0$ when simplifying $\frac{5a}{a^2} = \frac{5}{a}$?

Excellent observation. In the context of an ellipse, $a$ represents the semi-major axis length. A length of zero would imply the ellipse collapses into a point, which contradicts the definition of an ellipse. Therefore, $a$ must be non-zero, allowing us to safely divide by $a$.

Select Answer:

Visualized Solution

Problem Setup

Given Ellipse: $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$
Length of Latus Rectum ( $L R$ ) = $10$
Eccentricity ( $e$ ) = Minimum value of $f (t) = t^{2} + t + \frac{11}{12}$

Analyzing the Function $f (t)$

Function: $f (t) = t^{2} + t + \frac{11}{12}$
This is a quadratic function of the form $a t^{2} + b t + c$
Since $a = 1 > 0$ , the function has a minimum value at its vertex.

Finding the Vertex

Minimum occurs at $t = - \frac{b}{2 a}$
For $f (t) = 1 \cdot t^{2} + 1 \cdot t + \frac{11}{12}$
$t = - \frac{1}{2 ( 1 )} = - \frac{1}{2}$

Calculating the Minimum Value

Substitute $t = - \frac{1}{2}$ into $f (t)$ :
$f (- \frac{1}{2}) = (- \frac{1}{2})^{2} + (- \frac{1}{2}) + \frac{11}{12}$
$f_{min} = \frac{1}{4} - \frac{1}{2} + \frac{11}{12}$

Simplifying $f_{min}$

Common denominator is $12$ :
$f_{min} = \frac{3 - 6 + 11}{12}$
$f_{min} = \frac{8}{12} = \frac{2}{3}$
Therefore, eccentricity $e = \frac{2}{3}$

Latus Rectum Formula

Length of Latus Rectum for $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ is $\frac{2 b ^{2}}{a}$
Given: $L R = 10$
$\frac{2 b ^{2}}{a} = 10$

Relation between $a$ and $b$

$\frac{2 b ^{2}}{a} = 10$
$2 b^{2} = 10 a$
$b^{2} = 5 a$

Eccentricity Formula

Standard relation: $b^{2} = a^{2} (1 - e^{2})$
Alternatively: $e^{2} = 1 - \frac{b ^{2}}{a ^{2}}$

Substituting Known Values

Substitute $e = \frac{2}{3}$ and $b^{2} = 5 a$ :
$(\frac{2}{3})^{2} = 1 - \frac{5 a}{a ^{2}}$

Simplifying the Equation

$(\frac{2}{3})^{2} = \frac{4}{9}$
$\frac{5 a}{a ^{2}} = \frac{5}{a}$ (since $a \neq = 0$ )
$\frac{4}{9} = 1 - \frac{5}{a}$

Solving for $a$

Rearrange: $\frac{5}{a} = 1 - \frac{4}{9}$
$\frac{5}{a} = \frac{9 - 4}{9} = \frac{5}{9}$
Therefore, $a = 9$

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

We know $b^{2} = 5 a$
$b^{2} = 5 (9) = 45$
$a^{2} = 9^{2} = 81$

Final Calculation

We need to find $a^{2} + b^{2}$
$a^{2} + b^{2} = 81 + 45$
$a^{2} + b^{2} = 126$

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

Solution Diagram

Analyzing the Quadratic Function

To begin, we must determine the eccentricity

e

of the ellipse by finding the minimum value of the function

f (t) = t^{2} + t + \frac{11}{12}

. Since the coefficient of

t^{2}

is positive, the parabola opens upwards, and the minimum occurs at the vertex.

Using the vertex formula

t = - \frac{b}{2 a}

, we substitute the coefficients from

f (t)

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

Now, we substitute

t = - \frac{1}{2}

back into

f (t)

to find the minimum value:

f (- \frac{1}{2}) = (- \frac{1}{2})^{2} + (- \frac{1}{2}) + \frac{11}{12}

Simplifying the expression:

f (- \frac{1}{2}) = \frac{1}{4} - \frac{1}{2} + \frac{11}{12} = \frac{3 - 6 + 11}{12} = \frac{8}{12} = \frac{2}{3}

Thus, the eccentricity of the ellipse is

e = \frac{2}{3}

Geometric Constraints of the Ellipse

We are given that the length of the latus rectum of the ellipse

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

10

. The formula for the length of the latus rectum is

\frac{2 b ^{2}}{a}

Setting this equal to the given value:

\frac{2 b ^{2}}{a} = 10 \Rightarrow b^{2} = 5 a

We utilize the fundamental relationship between eccentricity and the semi-axes of an ellipse:

e^{2} = 1 - \frac{b ^{2}}{a ^{2}}

Solving for the Ellipse Parameters

Substituting

e = \frac{2}{3}

and

b^{2} = 5 a

into the eccentricity equation:

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

This simplifies to:

\frac{4}{9} = 1 - \frac{5}{a}

Rearranging the terms to solve for

a

\frac{5}{a} = 1 - \frac{4}{9} = \frac{5}{9}

From this, we find

a = 9

. Consequently, we calculate

b^{2}

b^{2} = 5 (9) = 45

Final Calculation

We are tasked with finding the value of

a^{2} + b^{2}

. Given

a = 9

, then

a^{2} = 81

Adding the values together:

a^{2} + b^{2} = 81 + 45 = 126

The final result is 126.

Common Traps & FAQs

Wait, why do we use the vertex formula

t = - \frac{b}{2 a}

? Can't we just guess the value?

Guessing is the enemy of precision! The function

f (t) = t^{2} + t + \frac{11}{12}

is a parabola opening upwards. Its minimum is strictly defined at the vertex. Using the formula

t = - \frac{b}{2 a}

is the mathematically rigorous way to ensure you find the absolute minimum, not just an approximation.

Why is the latus rectum formula

\frac{2 b ^{2}}{a}

and not

\frac{2 a ^{2}}{b}

Great question! In the standard ellipse equation

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

, we assume

a > b

. The latus rectum is the chord passing through the focus perpendicular to the major axis. The formula

\frac{2 b ^{2}}{a}

is specific to the major axis being along the x-axis. Always check your orientation!

Why do we reject the possibility of

a = 0

when simplifying

\frac{5 a}{a ^{2}} = \frac{5}{a}

Excellent observation. In the context of an ellipse,

a

represents the semi-major axis length. A length of zero would imply the ellipse collapses into a point, which contradicts the definition of an ellipse. Therefore,

a

must be non-zero, allowing us to safely divide by

a

Animated Solution for Mathematics - Conic Sections: Let the length of a latus rectum of an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ be 10 . If its eccentricity is the minimum value of the function $f (t) = t^{2} + t + \frac{11}{12}, t \in R$ , then $a^{2} + b^{2}$ is equal to :

Select Answer:

Visualized Solution

Problem Setup

Analyzing the Function $f (t)$

Finding the Vertex

Calculating the Minimum Value

Simplifying $f_{min}$

Latus Rectum Formula

Relation between $a$ and $b$

Eccentricity Formula

Substituting Known Values

Simplifying the Equation

Solving for $a$

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

Final Calculation

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

Analyzing the Quadratic Function

Geometric Constraints of the Ellipse

Solving for the Ellipse Parameters

Final Calculation

Similar Questions

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

Let $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b$ be an ellipse, whose eccentricity is $\frac{1}{2}$ and the length of the latus rectum is $14$ . Then the square of the eccentricity of $\frac{x ^{2}}{a ^{2}} - \frac{m y ^{2}}{b ^{2}} = 1$ is :

For some $θ \in (0, \frac{π}{2})$ , if the eccentricity of the hyperbola, $x^{2} - y^{2} sec^{2} θ = 10$ is $5$ times the eccentricity of the ellipse, $x^{2} sec^{2} θ + y^{2} = 5$ , then the length of the latus rectum of the ellipse, is :

In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is 10 and one of the foci is at $(0, 53)$ , then the length of its latus rectum is:

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An ellipse has its center at $(1, - 2)$ , one focus at $(3, - 2)$ and one vertex at $(5, - 2)$ . Then the length of its latus rectum is :

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Let the eccentricity of an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ is reciprocal to that of the hyperbola $2 x^{2} - 2 y^{2} = 1$ . If the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is _____.

Let $P$ be a parabola with vertex $(2, 3)$ and directrix $2 x + y = 6$ . Let an ellipse $E : \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b$ of eccentricity $\frac{1}{2}$ pass through the focus of the parabola $P$ . Then the square of the length of the latus rectum of $E$ , is

Select Answer:

Visualized Solution

Problem Setup

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

Finding the Vertex

Calculating the Minimum Value

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } fmin​

Latus Rectum Formula

Relation between .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; } b

Eccentricity Formula

Substituting Known Values

Simplifying the Equation

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

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

Final Calculation

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

Analyzing the Quadratic Function

Geometric Constraints of the Ellipse

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; } An ellipse has its center at (1,−2), one focus at (3,−2) and one vertex at (5,−2). Then the length of its latus rectum is :

Analyzing the Function $f (t)$

Simplifying $f_{min}$

Relation between $a$ and $b$

Solving for $a$

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

An ellipse has its center at $(1, - 2)$ , one focus at $(3, - 2)$ and one vertex at $(5, - 2)$ . Then the length of its latus rectum is :