JEE Advanced 2012

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Animated Solution for Mathematics - Conic Sections: The ellipse $E_{1} : \frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ is inscribed in a rectangle $R$ whose sides are parallel to the coordinate axes. Another ellipse $E_{2}$ passing through the point $(0, 4)$ circumscribes the rectangle $R$ . The eccentricity of the ellipse $E_{2}$ is

Q: Wait, why are the vertices of the rectangle $(\pm 3, \pm 2)$? I thought the ellipse vertices were just $(\pm 3, 0)$ and $(0, \pm 2)$.

Great catch! The ellipse $E_1$ has vertices at $(\pm 3, 0)$ and $(0, \pm 2)$. The rectangle $R$ is *inscribed around* the ellipse, meaning its sides are tangent to the ellipse. The horizontal sides are tangent at $(0, \pm 2)$, so they are the lines $y = 2$ and $y = -2$. The vertical sides are tangent at $(\pm 3, 0)$, so they are the lines $x = 3$ and $x = -3$. The intersection of these lines gives the rectangle's corners: $(\pm 3, \pm 2)$.

Q: How do I know if the ellipse is vertical or horizontal? Does it matter for the eccentricity formula?

It absolutely matters! The eccentricity formula $e = \sqrt{1 - \frac{\text{minor}^2}{\text{major}^2}}$ depends on which axis is longer. If $b^2 > a^2$, the major axis is vertical, so $e = \sqrt{1 - \frac{a^2}{b^2}}$. If $a^2 > b^2$, the major axis is horizontal, so $e = \sqrt{1 - \frac{b^2}{a^2}}$. Always compare $a^2$ and $b^2$ before calculating $e$!

Select Answer:

Visualized Solution

Visualizing the Setup

Ellipse $E_{1} : \frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$
Semi-major axis $a_{1} = 3$ , Semi-minor axis $b_{1} = 2$
Rectangle $R$ is inscribed around $E_{1}$ with sides parallel to the axes.

Vertices of Rectangle $R$

Sides of $R$ are tangent to $E_{1}$ at its vertices.
Horizontal tangents: $y = \pm 2$
Vertical tangents: $x = \pm 3$
Vertices of $R$ are $(\pm 3, \pm 2)$

Equation of Ellipse $E_{2}$

Let $E_{2} : \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$
Given: $E_{2}$ passes through $(0, 4)$

Finding Parameter $b^{2}$

Substitute $(0, 4)$ into $E_{2}$ :
$\frac{0 ^{2}}{a ^{2}} + \frac{4 ^{2}}{b ^{2}} = 1$
$\frac{16}{b ^{2}} = 1 ⟹ b^{2} = 16$

Substituting Rectangle Vertices

$E_{2}$ circumscribes $R ⟹$ passes through $(\pm 3, \pm 2)$
Substitute $(3, 2)$ and $b^{2} = 16$ into $E_{2}$

Solving for Parameter $a^{2}$

$\frac{3 ^{2}}{a ^{2}} + \frac{2 ^{2}}{16} = 1$
$\frac{9}{a ^{2}} + \frac{1}{4} = 1$

Finalizing $a^{2}$ Value

$\frac{9}{a ^{2}} = 1 - \frac{1}{4} = \frac{3}{4}$
$3 a^{2} = 36 ⟹ a^{2} = 12$

Analyzing Ellipse Orientation

$a^{2} = 12$ and $b^{2} = 16$
Since $b^{2} > a^{2}$ , $E_{2}$ is a vertical ellipse.

Calculating Eccentricity $e$

For vertical ellipse ( $b > a$ ):
$e = 1 - \frac{a ^{2}}{b ^{2}}$
$e = 1 - \frac{12}{16} = 1 - \frac{3}{4} = \frac{1}{4} = \frac{1}{2}$

The Way Forward

Key Takeaway: For vertical ellipses ( $b > a$ ), eccentricity is $e = 1 - a^{2} / b^{2}$ .
Next Challenge: How would the eccentricity change if $E_{2}$ passed through $(5, 0)$ instead of $(0, 4)$ ?

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

Solution Diagram

Analyzing the Setup

Welcome, fellow traveler in the world of conic sections! Today, we are choreographing a dance between two ellipses and a rectangle.

In the center, we have a stable ellipse,

E_{1}

, defined by the equation:

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

Its semi-major axis is

a_{1} = 3

and its semi-minor axis is

b_{1} = 2

A rectangle

R

perfectly hugs this ellipse with sides parallel to the coordinate axes. Its vertices are the points where the horizontal tangents

y = \pm 2

meet the vertical tangents

x = \pm 3

Thus, the corners of our rectangle are

(\pm 3, \pm 2)

The Mystery of the Second Ellipse

We introduce a second ellipse,

E_{2}

, which circumscribes our rectangle

R

. This means all four corners of the rectangle,

(\pm 3, \pm 2)

, must lie on the boundary of

E_{2}

We also know

E_{2}

passes through the point

(0, 4)

. Let the equation of

E_{2}

be:

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

Our mission is to find the eccentricity of

E_{2}

. To do this, we must determine the values of

a^{2}

and

b^{2}

The Algebraic Breakthrough

Since

E_{2}

passes through

(0, 4)

, we substitute these coordinates into the equation:

\frac{0 ^{2}}{a ^{2}} + \frac{4 ^{2}}{b ^{2}} = 1

The

x

-term vanishes, leaving us with

\frac{16}{b ^{2}} = 1

, which immediately tells us that

b^{2} = 16

Now, we use the fact that the rectangle's corner

(3, 2)

lies on

E_{2}

. Substituting

x = 3

y = 2

, and

b^{2} = 16

into the equation, we get:

\frac{3 ^{2}}{a ^{2}} + \frac{2 ^{2}}{16} = 1

This simplifies to

\frac{9}{a ^{2}} + \frac{4}{16} = 1

, or

\frac{9}{a ^{2}} + \frac{1}{4} = 1

Subtracting

\frac{1}{4}

from both sides, we find

\frac{9}{a ^{2}} = \frac{3}{4}

. Cross-multiplying gives

3 a^{2} = 36

, so

a^{2} = 12

Final Calculation

We have our parameters:

a^{2} = 12

and

b^{2} = 16

. Since

b^{2} > a^{2}

, our ellipse

E_{2}

is stretched vertically along the

y

-axis.

For a vertical ellipse, the eccentricity

e

is calculated using the formula:

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

Substituting our values, we get:

e = 1 - \frac{12}{16} = 1 - \frac{3}{4} = \frac{1}{4}

The final eccentricity is $e = \frac{1}{2}$ .

Common Traps & FAQs

Wait, why are the vertices of the rectangle

(\pm 3, \pm 2)

? I thought the ellipse vertices were just

(\pm 3, 0)

and

(0, \pm 2)

Great catch! The ellipse

E_{1}

has vertices at

(\pm 3, 0)

and

(0, \pm 2)

. The rectangle

R

is *inscribed around* the ellipse, meaning its sides are tangent to the ellipse. The horizontal sides are tangent at

(0, \pm 2)

, so they are the lines

y = 2

and

y = - 2

. The vertical sides are tangent at

(\pm 3, 0)

, so they are the lines

x = 3

and

x = - 3

. The intersection of these lines gives the rectangle's corners:

(\pm 3, \pm 2)

Why does

E_{2}

pass through

(0, 4)

? Is that just a random point?

It is not random at all! It is a boundary condition provided by the problem to fix the size of the ellipse

E_{2}

. Without knowing that

E_{2}

passes through

(0, 4)

, we would have infinitely many ellipses that could circumscribe the rectangle. This point allows us to solve for the vertical semi-axis

b

immediately.

How do I know if the ellipse is vertical or horizontal? Does it matter for the eccentricity formula?

It absolutely matters! The eccentricity formula

e = 1 - \frac{minor ^{2}}{major ^{2}}

depends on which axis is longer. If

b^{2} > a^{2}

, the major axis is vertical, so

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

. If

a^{2} > b^{2}

, the major axis is horizontal, so

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

. Always compare

a^{2}

and

b^{2}

before calculating

e

Select Answer:

Visualized Solution

Visualizing the Setup

Vertices of Rectangle $R$

Equation of Ellipse $E_{2}$

Finding Parameter $b^{2}$

Substituting Rectangle Vertices

Solving for Parameter $a^{2}$

Finalizing $a^{2}$ Value

Analyzing Ellipse Orientation

Calculating Eccentricity $e$

The Way Forward

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

Analyzing the Setup

The Mystery of the Second Ellipse

The Algebraic Breakthrough

Final Calculation

Similar Questions

The ellipse $x^{2} + 4 y^{2} = 4$ is inscribed in a rectangle aligned with the coordinate axes, which in turn is inscribed in another ellipse that passes through the point $(4, 0)$ . Then the equation of the ellipse is:

Let the maximum area of the triangle that can be inscribed in the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{4} = 1, a > 2$ , having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the y-axis, be $63$ . Then the eccentricity of the ellipse is :

Select Answer:

Visualized Solution

Visualizing the Setup

Vertices of Rectangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

Equation of Ellipse .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E2​

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

Substituting Rectangle Vertices

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

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

Analyzing Ellipse Orientation

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

The Way Forward

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

Analyzing the Setup

The Mystery of the Second Ellipse

The Algebraic Breakthrough

Final Calculation

Similar Questions

Vertices of Rectangle $R$

Equation of Ellipse $E_{2}$

Finding Parameter $b^{2}$

Solving for Parameter $a^{2}$

Finalizing $a^{2}$ Value

Calculating Eccentricity $e$