JEE Main 2019 (12 January)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: Let $S$ and $S^{'}$ be the foci of the ellipse and $B$ be any one of the extremities of its minor axis. If $S^{'} B S$ is a right angled triangle with right angle at $B$ and area $(S^{'} B S) = 8$ sq. units, then the length of a latus rectum of the ellipse is :

Select Answer:

Progressive Hints

Hint 1

Visualize the ellipse on a Cartesian plane. The triangle

S^{'} B S

is formed by the two foci

S (a e, 0)

and

S^{'} (- a e, 0)

and the extremity of the minor axis

B (0, b)

. The right angle at

B

implies that the line segments

S B

and

S^{'} B

are perpendicular, which gives you a direct algebraic relationship between

a

e

, and

b

Hint 2

Recall the coordinate geometry condition for perpendicular lines: the product of their slopes must be

- 1

. Calculate the slope of

S B

and

S^{'} B

using the coordinates

S (a e, 0)

S^{'} (- a e, 0)

, and

B (0, b)

. This will lead you to the elegant result

b^{2} = a^{2} e^{2}

Hint 3

Use the area of the triangle

△ S^{'} B S = 8

to find another equation. The base of the triangle is the distance between the foci,

2 a e

, and the height is

b

. Combine the area equation

\frac{1}{2} \times (2 a e) \times b = 8

with the relation

b = a e

to find the value of

b

. Finally, use the fundamental ellipse identity

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

to solve for

a

and calculate the latus rectum

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

Visualized Solution

Standard Ellipse and Foci

Let the equation of the ellipse be $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ where $a > b$ .
The foci are located at $S (a e, 0)$ and $S^{'} (- a e, 0)$ .

Defining Point $B$ and $△ S^{'} B S$

$B$ is an extremity of the minor axis, so its coordinates are $B (0, b)$ .
Connect $S^{'}$ , $B$ , and $S$ to form $△ S^{'} B S$ .

The Right Angle Condition

The problem states that $△ S^{'} B S$ is right-angled at $B$ .
This means the line segments $S^{'} B$ and $S B$ are perpendicular.
Therefore, the product of their slopes must be $- 1$ : $m_{S^{'} B} \times m_{S B} = - 1$ .

Calculating the Slopes

Slope of $S B$ : $m_{S B} = \frac{b - 0}{0 - a e} = - \frac{b}{a e}$
Slope of $S^{'} B$ : $m_{S^{'} B} = \frac{b - 0}{0 - ( - a e )} = \frac{b}{a e}$
Setting the product to $- 1$ : $(\frac{b}{a e}) \times (- \frac{b}{a e}) = - 1$

Simplifying the Slope Equation

Multiply the terms: $- \frac{b ^{2}}{a ^{2} e ^{2}} = - 1$
Cancel the negative signs: $\frac{b ^{2}}{a ^{2} e ^{2}} = 1$
Cross-multiply to get: $b^{2} = a^{2} e^{2}$
Taking the square root (since lengths are positive): $b = a e$

The Area Condition

The second piece of given information is the area of $△ S^{'} B S$ .
Area of $△ S^{'} B S = 8$ sq. units.
The formula for the area of a triangle is $\frac{1}{2} \times base \times height$ .

Setting up the Area Equation

The base of the triangle is the distance between the foci, $S S^{'} = 2 a e$ .
The height of the triangle is the distance from the origin to $B$ , which is $b$ .
Substituting into the formula: $\frac{1}{2} \times (2 a e) \times b = 8$

Solving for $b$

Simplify the left side: $a e b = 8$
From our earlier result, we know $a e = b$ .
Substitute $a e$ with $b$ : $b \times b = 8 ⟹ b^{2} = 8$
Therefore, $b = 22$

Relating $a$ , $b$ , and $e$

We have the value of $b^{2}$ . To find the latus rectum, we also need the value of $a$ .
Recall the fundamental relation for an ellipse: $b^{2} = a^{2} (1 - e^{2})$

Substituting Known Values

Expand the relation: $b^{2} = a^{2} - a^{2} e^{2}$
We know that $a^{2} e^{2} = (a e)^{2}$ .
Since $a e = b$ , we can substitute $a^{2} e^{2}$ with $b^{2}$ .
This gives: $b^{2} = a^{2} - b^{2}$

Solving for $a$

Rearrange the equation: $a^{2} = b^{2} + b^{2} = 2 b^{2}$
We already found that $b^{2} = 8$ .
Substitute $b^{2} = 8$ : $a^{2} = 2 \times 8 = 16$
Taking the square root: $a = 4$

Formula for Latus Rectum

We now have $a = 4$ and $b^{2} = 8$ .
The formula for the length of the latus rectum of an ellipse is $\frac{2 b ^{2}}{a}$ .

Substituting into Latus Rectum Formula

Length of Latus Rectum $= \frac{2 b ^{2}}{a}$
Substitute $b^{2} = 8$ and $a = 4$ .
Length $= \frac{2 \times 8}{4}$

Final Calculation

Length $= \frac{16}{4} = 4$
The length of the latus rectum is $4$ units.
This matches option (3).

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

Solution Diagram

Analyzing the Setup

Imagine you are standing on the coordinate plane, looking at a perfectly symmetrical ellipse. It is a shape defined by its constraints, and in this problem, those constraints are beautifully intertwined.

We are given an ellipse with foci

S (a e, 0)

and

S^{'} (- a e, 0)

and an extremity of the minor axis

B (0, b)

. When we connect these points to form

△ S^{'} B S

, we are defining the very soul of the ellipse.

The problem states that

△ S^{'} B S

is a right-angled triangle at

B

. Geometrically, this means the line segments

S B

and

S^{'} B

are perpendicular.

In the language of coordinate geometry, this translates to the product of their slopes being

- 1

. Let us calculate these slopes:

S l o p e_{S B} = \frac{b - 0}{0 - a e} = - \frac{b}{a e}

S l o p e_{S^{'} B} = \frac{b - 0}{0 - ( - a e )} = \frac{b}{a e}

Multiplying these gives us:

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

This is a profound realization: for this specific ellipse, the distance from the center to the focus is exactly equal to the semi-minor axis.

The Area Constraint

Unlocking the Values

Now that we have established the geometric relationship, we turn to the area. We are told that the area of

△ S^{'} B S

8

square units.

Using the standard formula for the area of a triangle,

Area = \frac{1}{2} \times base \times height

, we identify the base as the distance between the foci,

S S^{'} = 2 a e

, and the height as the distance from the origin to

B

, which is

b

Substituting these, we get:

\frac{1}{2} \times (2 a e) \times b = 8 \Rightarrow a e b = 8

Since we already know

a e = b

, we can substitute

b

for

a e

to get:

b \times b = 8 \Rightarrow b^{2} = 8

This gives us

b = 22

. We are halfway there!

The Final Synthesis

We have

b^{2} = 8

, but we need the length of the latus rectum, which is defined as

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

. We are currently missing

a

To find it, we use the fundamental eccentricity relation for an ellipse:

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

. Expanding this, we get:

b^{2} = a^{2} - a^{2} e^{2}

Since

a^{2} e^{2} = (a e)^{2}

and we know

a e = b

, this becomes:

b^{2} = a^{2} - b^{2} \Rightarrow a^{2} = 2 b^{2}

Substituting

b^{2} = 8

, we get

a^{2} = 2 \times 8 = 16

, so

a = 4

Now, we have all the pieces of our puzzle:

a = 4

and

b^{2} = 8

. The length of the latus rectum is:

\frac{2 b ^{2}}{a} = \frac{2 \times 8}{4} = \frac{16}{4} = 4

The final length of the latus rectum is 4 units.

Common Traps & FAQs

Wait, why is the base of the triangle

2 a e

? Isn't the base just the distance between the foci?

You are absolutely correct! The base of the triangle is indeed the distance between the two foci,

S

and

S^{'}

. Since the foci are at

(a e, 0)

and

(- a e, 0)

, the distance between them is

∣ a e - (- a e) ∣ = 2 a e

. Because the triangle is symmetric about the y-axis, the height is simply the y-coordinate of point

B

, which is

b

. So, the area formula

\frac{1}{2} \times base \times height

becomes

\frac{1}{2} \times (2 a e) \times b = a e b

Why do we set the product of the slopes to

- 1

That is the fundamental condition for two lines to be perpendicular in the Cartesian plane. If a line has slope

m_{1}

and another has slope

m_{2}

, they are perpendicular if and only if

m_{1} \times m_{2} = - 1

. In this problem, the right angle at

B

tells us that

S B ⊥ S^{'} B

, so this is the most direct way to translate that geometric fact into an algebraic equation.

Can we use the property that the sum of focal distances

S B + S^{'} B = 2 a

That is a brilliant observation! You absolutely can. In a right-angled triangle

S^{'} B S

, by the Pythagorean theorem,

S B^{2} + S^{'} B^{2} = (S^{'} S)^{2}

. Since the triangle is isosceles (because

B

is on the y-axis and the foci are symmetric),

S B = S^{'} B

. Thus,

2 S B^{2} = (2 a e)^{2}

, which simplifies to

S B^{2} = 2 a^{2} e^{2}

. This is a powerful alternative path to the same result. Keep looking for these connections!

Select Answer:

Visualized Solution

Standard Ellipse and Foci

Defining Point $B$ and $△ S^{'} B S$

The Right Angle Condition

Calculating the Slopes

Simplifying the Slope Equation

The Area Condition

Setting up the Area Equation

Solving for $b$

Relating $a$ , $b$ , and $e$

Substituting Known Values

Solving for $a$

Formula for Latus Rectum

Substituting into Latus Rectum Formula

Final Calculation

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

Analyzing the Setup

The Area Constraint

The Final Synthesis

Similar Questions

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 _____.

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Let the foci of a hyperbola coincide with the foci of the ellipse $\frac{x ^{2}}{36} + \frac{y ^{2}}{16} = 1$ . If the eccentricity of the hyperbola is 5, then the length of its latus rectum is :

Select Answer:

Visualized Solution

Standard Ellipse and Foci

Defining Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △S′BS

The Right Angle Condition

Calculating the Slopes

Simplifying the Slope Equation

The Area Condition

Setting up the Area Equation

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

Substituting Known Values

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

Formula for Latus Rectum

Substituting into Latus Rectum Formula

Final Calculation

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

Analyzing the Setup

The Area Constraint

The Final Synthesis

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