JEE Main 2015

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1$ , is:

Q: Wait, why can't we just use the area of the ellipse formula?

The area of the ellipse is $\pi ab$. But the question asks for the area of the quadrilateral formed by the tangents at the endpoints of the latera recta, which is a completely different geometric shape—a rhombus! We must calculate the area of this specific quadrilateral.

Q: Is there a shortcut formula?

Yes, the area is given by $\frac{2a^2}{e}$. However, in the JEE, understanding the derivation is crucial for solving variations of this problem. Always prioritize the geometric derivation first!

Select Answer:

Visualized Solution

Standard Equation of the Ellipse

Given ellipse: $\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1$
Comparing with standard form $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ :
We get $a^{2} = 9 ⟹ a = 3$ (semi-major axis)
And $b^{2} = 5 ⟹ b = 5$ (semi-minor axis)

Finding the Eccentricity $e$

Eccentricity formula: $e = 1 - \frac{b ^{2}}{a ^{2}}$
Substituting $a^{2} = 9$ and $b^{2} = 5$ :
$e = 1 - \frac{5}{9} = \frac{4}{9} = \frac{2}{3}$

Endpoints of the Latera Recta

The focus is at $(\pm a e, 0) = (\pm 2, 0)$
The endpoints of the latera recta are given by $(\pm a e, \pm \frac{b ^{2}}{a})$
Substituting values: $a e = 3 \times \frac{2}{3} = 2$ and $\frac{b ^{2}}{a} = \frac{5}{3}$
Endpoint in the first quadrant: $L (2, \frac{5}{3})$

Tangent Equation at $L (2, \frac{5}{3})$

Equation of a tangent at $(x_{1}, y_{1})$ : $\frac{x x _{1}}{a ^{2}} + \frac{y y _{1}}{b ^{2}} = 1$
Substituting $x_{1} = 2$ and $y_{1} = \frac{5}{3}$ :
$\frac{x ( 2 )}{9} + \frac{y ( \frac{5}{3} )}{5} = 1$

Simplifying the Tangent Equation

The equation is: $\frac{2 x}{9} + \frac{y ( \frac{5}{3} )}{5} = 1$
Simplifying the second term: $\frac{5 y}{3 \times 5} = \frac{y}{3}$
The simplified equation of the tangent is:
$\frac{2 x}{9} + \frac{y}{3} = 1$

Intercepts on the Axes

To find the x-intercept, set $y = 0$ :
$\frac{2 x}{9} = 1 ⟹ x = \frac{9}{2}$
So, the x-intercept is $A (\frac{9}{2}, 0)$
To find the y-intercept, set $x = 0$ :
$\frac{y}{3} = 1 ⟹ y = 3$
So, the y-intercept is $B (0, 3)$

Area of Triangle in First Quadrant

The tangent line forms a right-angled triangle $O A B$ with the coordinate axes.
Area of $Δ O A B = \frac{1}{2} \times base \times height$
Base $O A = \frac{9}{2}$ and Height $O B = 3$
Area $= \frac{1}{2} \times \frac{9}{2} \times 3 = \frac{27}{4}$ sq. units

Exploiting Symmetry

The ellipse is symmetric about both the x-axis and the y-axis.
The four endpoints of the latera recta are symmetric across all four quadrants.
Therefore, the four tangents form a symmetric quadrilateral (rhombus).
The total area is exactly 4 times the area of the triangle $O A B$ in the first quadrant.

Calculating the Total Area

Total Area $= 4 \times Area (Δ O A B)$
Total Area $= 4 \times \frac{27}{4}$
Total Area $= 27$ sq. units

Key Takeaway & General Formula

For any ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ , the area of the quadrilateral formed by the tangents at the endpoints of the latera recta is:
$Area = \frac{2 a ^{2}}{e}$
Let's verify: $\frac{2 ( 9 )}{2/3} = 18 \times \frac{3}{2} = 27$
Thus, the correct option is 27 (Option 2).

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The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Welcome, future engineer! Today, we are going to unravel the geometry of the ellipse defined by the equation:

\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1

Comparing this to the standard form

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

, we identify the parameters:

a^{2} = 9 \Rightarrow a = 3

b^{2} = 5 \Rightarrow b = 5

The Latus Rectum

The latus rectum is the chord passing through the focus, perpendicular to the major axis. To find it, we first calculate the eccentricity

e

using the formula:

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

Substituting our values, we obtain:

e = 1 - \frac{5}{9} = \frac{4}{9} = \frac{2}{3}

The foci are located at

(\pm a e, 0)

, which simplifies to

(\pm 2, 0)

. The endpoints of the latus rectum are given by

(\pm a e, \pm \frac{b ^{2}}{a})

Plugging in our values, the endpoint in the first quadrant is

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

The Tangent

Next, we determine the equation of the tangent at point

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

. The formula for a tangent at

(x_{1}, y_{1})

is:

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

Substituting

x_{1} = 2

and

y_{1} = \frac{5}{3}

, we get:

\frac{x ( 2 )}{9} + \frac{y ( \frac{5}{3} )}{5} = 1

Simplifying this, the

5

s cancel out, resulting in the equation of the tangent:

\frac{2 x}{9} + \frac{y}{3} = 1

Final Calculation

To find the area, we determine where this line intersects the coordinate axes. Setting

y = 0

, we find the

x

-intercept:

\frac{2 x}{9} = 1 \Rightarrow x = \frac{9}{2}

Setting

x = 0

, we find the

y

-intercept:

\frac{y}{3} = 1 \Rightarrow y = 3

This tangent forms a right-angled triangle with the axes, having a base of

\frac{9}{2}

and a height of

3

. The area of this triangle is:

Area = \frac{1}{2} \times \frac{9}{2} \times 3 = \frac{27}{4}

Due to the symmetry of the ellipse, there are four such triangles, one in each quadrant. The total area is:

4 \times \frac{27}{4} = 27

The final area is 27 square units.

Common Traps & FAQs

Wait, why can't we just use the area of the ellipse formula?

The area of the ellipse is

π ab

. But the question asks for the area of the quadrilateral formed by the tangents at the endpoints of the latera recta, which is a completely different geometric shape—a rhombus! We must calculate the area of this specific quadrilateral.

How do I know the tangents form a rhombus?

Because the ellipse is symmetric about both axes, the four tangents at the four endpoints of the latera recta are reflections of each other across the axes, creating a perfectly symmetric quadrilateral with equal sides and diagonals, which is a rhombus.

Is there a shortcut formula?

Yes, the area is given by

\frac{2 a ^{2}}{e}

. However, in the JEE, understanding the derivation is crucial for solving variations of this problem. Always prioritize the geometric derivation first!

Animated Solution for Mathematics - Conic Sections: The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1$ , is:

Select Answer:

Visualized Solution

Standard Equation of the Ellipse

Finding the Eccentricity $e$

Endpoints of the Latera Recta

Tangent Equation at $L (2, \frac{5}{3})$

Simplifying the Tangent Equation

Intercepts on the Axes

Area of Triangle in First Quadrant

Exploiting Symmetry

Calculating the Total Area

Key Takeaway & General Formula

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Latus Rectum

The Tangent

Final Calculation

Similar Questions

The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1$ , is

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On the ellipse $4 x^{2} + 9 y^{2} = 1$ , the points at which the tangents are parallel to the line $8 x = 9 y$ are

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Select Answer:

Visualized Solution

Standard Equation of the Ellipse

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

Endpoints of the Latera Recta

Tangent Equation at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L(2,35​)

Simplifying the Tangent Equation

Intercepts on the Axes

Area of Triangle in First Quadrant

Exploiting Symmetry

Calculating the Total Area

Key Takeaway & General Formula

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Latus Rectum

The Tangent

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse 9x2​+5y2​=1, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } On the ellipse 4x2+9y2=1, the points at which the tangents are parallel to the line 8x=9y are

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The tangent and normal to the ellipse 3x2+5y2=32 at the point P(2,2) meet the x-axis at Q and R, respectively. Then the area (in sq. units) of the triangle PQR is :

Finding the Eccentricity $e$

Tangent Equation at $L (2, \frac{5}{3})$

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The tangent and normal to the ellipse $3 x^{2} + 5 y^{2} = 32$ at the point $P (2, 2)$ meet the x-axis at Q and R, respectively. Then the area (in sq. units) of the triangle PQR is :