JEE Main 2024 (27 Jan Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: Let $e_{1}$ be the eccentricity of the hyperbola $\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1$ and $e_{2}$ be the eccentricity of the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b$ , which passes through the foci of the hyperbola. If $e_{1} e_{2} = 1$ , then the length of the chord of the ellipse parallel to the x-axis and passing through $(0, 2)$ is :

Q: Why is the chord length just $2x$?

Because the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ is symmetric about both the $x$ and $y$ axes. When you intersect it with the horizontal line $y = 2$, the intersection points are $(\frac{5\sqrt{5}}{3}, 2)$ and $(-\frac{5\sqrt{5}}{3}, 2)$. The distance between them is simply the difference in their $x$-coordinates, which is $2 \times \frac{5\sqrt{5}}{3}$.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Visualize the geometric connection between the hyperbola and the ellipse. The hyperbola's foci are the key. Since the ellipse passes through these points and the major axis is horizontal, these foci are actually the vertices of your ellipse. This immediately gives you the value of the semi-major axis,

a

Hint 2

Hint 2: The Arsenal. You have the eccentricity of the hyperbola,

e_{1} = 1 + \frac{b ^{2}}{a ^{2}}

. Use the given condition

e_{1} e_{2} = 1

to find

e_{2}

. Then, apply the ellipse eccentricity formula

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

to find the missing parameter

b^{2}

for the ellipse.

Hint 3

Hint 3: The Execution. Once you have the full equation of the ellipse,

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

, substitute

y = 2

into it. This will give you the

x

-coordinates of the intersection points. The length of the chord is simply the difference between these two

x

-values.

Visualized Solution

Analyze the Hyperbola

Hyperbola $H : \frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1$
Standard form: $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$
Here, $a^{2} = 16$ and $b^{2} = 9$

Eccentricity of Hyperbola ( $e_{1}$ )

Formula: $e_{1} = 1 + \frac{b ^{2}}{a ^{2}}$
Substitute values: $e_{1} = 1 + \frac{9}{16}$
$e_{1} = \frac{25}{16} = \frac{5}{4}$

Foci of the Hyperbola

Foci coordinates: $(\pm a e_{1}, 0)$
We know $a = 4$ and $e_{1} = \frac{5}{4}$
Foci $= (\pm 4 \cdot \frac{5}{4}, 0) = (\pm 5, 0)$

Eccentricity of the Ellipse ( $e_{2}$ )

Given condition: $e_{1} e_{2} = 1$
Substitute $e_{1} = \frac{5}{4}$
$\frac{5}{4} \cdot e_{2} = 1 ⟹ e_{2} = \frac{4}{5}$

Ellipse Passing Through Foci

Ellipse $E : \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ (where $a > b$ )
It passes through the hyperbola's foci $(\pm 5, 0)$
Since these lie on the major axis, the semi-major axis $a = 5$

Finding Ellipse Parameter $b^{2}$

Ellipse eccentricity formula: $e_{2}^{2} = 1 - \frac{b ^{2}}{a ^{2}}$
Substitute $e_{2} = \frac{4}{5}$ and $a = 5$ :
$(\frac{4}{5})^{2} = 1 - \frac{b ^{2}}{25}$

Computing $b^{2}$

$\frac{16}{25} = 1 - \frac{b ^{2}}{25}$
$\frac{b ^{2}}{25} = 1 - \frac{16}{25} = \frac{9}{25}$
Therefore, $b^{2} = 9$

Equation of the Ellipse

Substitute $a^{2} = 25$ and $b^{2} = 9$
Equation of Ellipse $E : \frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$

Identifying the Chord

The chord is parallel to the x-axis.
It passes through the point $(0, 2)$ .
Therefore, the equation of the chord is the horizontal line $y = 2$ .

Finding Intersection Points

Substitute $y = 2$ into the ellipse equation:
$\frac{x ^{2}}{25} + \frac{2 ^{2}}{9} = 1 ⟹ \frac{x ^{2}}{25} + \frac{4}{9} = 1$
$\frac{x ^{2}}{25} = 1 - \frac{4}{9} = \frac{5}{9}$

Solving for x-coordinates

$x^{2} = 25 \cdot \frac{5}{9} = \frac{125}{9}$
$x = \pm \frac{125}{3} = \pm \frac{5 5}{3}$
The endpoints are $(\pm \frac{5 5}{3}, 2)$

Calculating Chord Length

Length is the distance between $(\frac{5 5}{3}, 2)$ and $(- \frac{5 5}{3}, 2)$
Length $= \frac{5 5}{3} - (- \frac{5 5}{3})$
Length $= \frac{10 5}{3}$

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The Sigma Insight: Foci, Directrices, and Eccentricity

Solution Diagram

Analyzing the Hyperbola

We begin with the hyperbola

H

defined by the equation:

\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1

By comparing this to the standard form

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

, we identify

a^{2} = 16

and

b^{2} = 9

, which gives

a = 4

and

b = 3

Next, we calculate the eccentricity

e_{1}

using the standard relation:

e_{1} = 1 + \frac{b ^{2}}{a ^{2}} = 1 + \frac{9}{16} = \frac{25}{16} = \frac{5}{4}

The foci of the hyperbola are located at

(\pm a e_{1}, 0)

. Substituting our values, we find the foci at

(\pm 5, 0)

The Ellipse Emerges

The problem states that the ellipse passes through these foci. Since the foci

(\pm 5, 0)

lie on the

x

-axis, the semi-major axis

a

of the ellipse is

5

We are given the condition

e_{1} e_{2} = 1

. Given

e_{1} = \frac{5}{4}

, it follows that

e_{2} = \frac{4}{5}

We use the eccentricity formula for an ellipse,

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

, to find the semi-minor axis

b

(\frac{4}{5})^{2} = 1 - \frac{b ^{2}}{5 ^{2}}

\frac{16}{25} = 1 - \frac{b ^{2}}{25} \Rightarrow \frac{b ^{2}}{25} = \frac{9}{25} \Rightarrow b^{2} = 9

Thus, the equation of the ellipse is:

\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1

The Final Calculation

We seek the length of a chord parallel to the

x

-axis passing through

(0, 2)

. This corresponds to the line

y = 2

Substituting

y = 2

into the ellipse equation, we obtain:

\frac{x ^{2}}{25} + \frac{4}{9} = 1

\frac{x ^{2}}{25} = 1 - \frac{4}{9} = \frac{5}{9}

Solving for

x

, we find:

x^{2} = \frac{125}{9} \Rightarrow x = \pm \frac{5 5}{3}

The chord connects the points

(- \frac{5 5}{3}, 2)

and

(\frac{5 5}{3}, 2)

. The length of this chord is the horizontal distance between these points:

Length = \frac{5 5}{3} - (- \frac{5 5}{3}) = \frac{10 5}{3}

Common Traps & FAQs

Wait, why do we assume the vertices of the ellipse are the foci of the hyperbola?

Great question! The problem states the ellipse passes through the foci of the hyperbola. Since the hyperbola's foci lie on the

x

-axis at

(\pm 5, 0)

and the ellipse's major axis is also on the

x

-axis (because

a > b

), these points must be the vertices of the ellipse. It is a beautiful geometric alignment!

Can I use the relation

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

for the ellipse?

Absolutely! That is the standard definition of eccentricity for an ellipse. Since we found

e_{2} = \frac{4}{5}

and

a = 5

, you can directly calculate

b^{2} = 25 (1 - \frac{16}{25}) = 25 (\frac{9}{25}) = 9

. It is the most efficient path forward.

Why is the chord length just

2 x

Because the ellipse

\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1

is symmetric about both the

x

and

y

axes. When you intersect it with the horizontal line

y = 2

, the intersection points are

(\frac{5 5}{3}, 2)

and

(- \frac{5 5}{3}, 2)

. The distance between them is simply the difference in their

x

-coordinates, which is

2 \times \frac{5 5}{3}

Select Answer:

Visualized Solution

Analyze the Hyperbola

Eccentricity of Hyperbola ( $e_{1}$ )

Foci of the Hyperbola

Eccentricity of the Ellipse ( $e_{2}$ )

Ellipse Passing Through Foci

Finding Ellipse Parameter $b^{2}$

Computing $b^{2}$

Equation of the Ellipse

Identifying the Chord

Finding Intersection Points

Solving for x-coordinates

Calculating Chord Length

The Sigma Insight: Foci, Directrices, and Eccentricity

Analyzing the Hyperbola

The Ellipse Emerges

The Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Analyze the Hyperbola

Eccentricity of Hyperbola ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e1​)

Foci of the Hyperbola

Eccentricity of the Ellipse ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e2​)

Ellipse Passing Through Foci

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

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

Equation of the Ellipse

Identifying the Chord

Finding Intersection Points

Solving for x-coordinates

Calculating Chord Length

The Sigma Insight: Foci, Directrices, and Eccentricity

Analyzing the Hyperbola

The Ellipse Emerges

The Final Calculation

Similar Questions

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