JEE Main 2021 (26 Aug Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: On the ellipse $\frac{x ^{2}}{8} + \frac{y ^{2}}{4} = 1$ let $P$ be a point in the second quadrant such that the tangent at $P$ to the ellipse is perpendicular to the line $x + 2 y = 0$ . Let $S$ and $S^{'}$ be the foci of the ellipse and $e$ be its eccentricity. If $A$ is the area of the triangle $S P S^{'}$ then, the value of $(5 - e^{2}) A$ is :

Q: Wait, why did we choose the positive y-intercept ($y = 2x + 6$) instead of the negative one ($y = 2x - 6$)?

Great observation! The problem explicitly places point $P$ in the second quadrant. In the second quadrant, $x$ is negative and $y$ is positive. If we chose the negative intercept, the line would pass through the fourth quadrant (where $y$ is negative). By choosing the positive intercept, we ensure the tangent line passes through the upper-left region of the coordinate plane, satisfying the condition.

Q: Is there a faster way to find the point of contact $P$ without using the formula $(-\frac{a^2m}{c}, \frac{b^2}{c})$?

Yes, you could use the parametric form $P(a \cos \theta, b \sin \theta)$ and equate the slope of the tangent at that point to $m=2$. However, the formula $(-\frac{a^2m}{c}, \frac{b^2}{c})$ is derived directly from the condition of tangency and is often much faster for competitive exams like JEE. It saves precious time and reduces the chance of algebraic errors.

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the ellipse. It is a horizontal ellipse with

a^{2} = 8

and

b^{2} = 4

. The tangent is perpendicular to

x + 2 y = 0

, which implies its slope is

m = 2

. Before diving into algebra, sketch the ellipse and the line. Where must the tangent touch to be in the second quadrant?

Hint 2

Hint 2: You need the condition of tangency for an ellipse:

y = m x \pm a^{2} m^{2} + b^{2}

. Use this to find the constant

c

in the tangent equation

y = m x + c

. Remember, the point of contact

(x_{1}, y_{1})

for a tangent

y = m x + c

is given by

(- \frac{a ^{2} m}{c}, \frac{b ^{2}}{c})

Hint 3

Hint 3: Once you have the coordinates of

P

, the triangle

S P S^{'}

has its base on the x-axis between the foci

S (2, 0)

and

S^{'} (- 2, 0)

. The base length is the distance between the foci, and the height is simply the y-coordinate of

P

. Calculate the area

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

and finish the calculation.

Visualized Solution

Ellipse Equation

Given Ellipse: $\frac{x ^{2}}{8} + \frac{y ^{2}}{4} = 1$
Standard form: $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$
Here, $a^{2} = 8$ and $b^{2} = 4$

Eccentricity Formula

Eccentricity formula: $e^{2} = 1 - \frac{b ^{2}}{a ^{2}}$
Substitute values: $e^{2} = 1 - \frac{4}{8}$

Calculating Eccentricity

$e^{2} = 1 - \frac{1}{2} = \frac{1}{2}$
Thus, $e = \frac{1}{2}$

Locating the Foci

Foci coordinates: $(\pm a e, 0)$
$a e = 8 \cdot \frac{1}{2} = 4 = 2$
Foci: $S (2, 0)$ and $S^{'} (- 2, 0)$

Slope of the Tangent

Given line: $x + 2 y = 0 \Rightarrow y = - \frac{1}{2} x$
Slope of given line $m_{L} = - \frac{1}{2}$
Tangent is perpendicular: $m \cdot m_{L} = - 1$
Slope of tangent $m = 2$

Equation of Tangent

Tangent equation: $y = m x \pm a^{2} m^{2} + b^{2}$
Substitute $m = 2, a^{2} = 8, b^{2} = 4$ :
$y = 2 x \pm 8 (2^{2}) + 4$

Simplifying Tangent Equation

$y = 2 x \pm 32 + 4$
$y = 2 x \pm 36$
$y = 2 x \pm 6$

Quadrant Constraint

Point $P$ is in the second quadrant
For a positive slope $m = 2$ , the tangent in the 2nd quadrant must have a positive y-intercept
Selected Tangent: $y = 2 x + 6$

Point of Contact Formula

Point of contact $P (x_{1}, y_{1}) = (- \frac{a ^{2} m}{c}, \frac{b ^{2}}{c})$
Here $c = 6, m = 2, a^{2} = 8, b^{2} = 4$

Calculating Point P

$x_{1} = - \frac{8 ( 2 )}{6} = - \frac{16}{6} = - \frac{8}{3}$
$y_{1} = \frac{4}{6} = \frac{2}{3}$
Point $P = (- \frac{8}{3}, \frac{2}{3})$

Triangle $Δ S P S^{'}$

Vertices of $Δ S P S^{'}$ : $S (2, 0)$ , $S^{'} (- 2, 0)$ , $P (- \frac{8}{3}, \frac{2}{3})$
Base $S S^{'} = 2 - (- 2) = 4$
Height $h = y$ -coordinate of $P = \frac{2}{3}$

Area of Triangle $A$

Area $A = \frac{1}{2} \times base \times height$
$A = \frac{1}{2} \times 4 \times \frac{2}{3}$
$A = 2 \times \frac{2}{3} = \frac{4}{3}$

Final Calculation

We need to find $(5 - e^{2}) A$
Substitute $e^{2} = \frac{1}{2}$ and $A = \frac{4}{3}$ :
$(5 - \frac{1}{2}) \times \frac{4}{3} = \frac{9}{2} \times \frac{4}{3}$
$= \frac{9 \times 4}{2 \times 3} = \frac{36}{6} = 6$

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

We begin with the equation of the ellipse:

\frac{x ^{2}}{8} + \frac{y ^{2}}{4} = 1

By comparing this to the standard form

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

, we identify the parameters

a^{2} = 8

and

b^{2} = 4

. This represents a horizontal ellipse stretched along the

x

-axis.

The eccentricity

e

is defined by the relationship:

e^{2} = 1 - \frac{b ^{2}}{a ^{2}} = 1 - \frac{4}{8} = \frac{1}{2}

Thus,

e = \frac{1}{2}

. The foci

S

and

S^{'}

are located at

(\pm a e, 0)

. Calculating

a e

, we find

8 \cdot \frac{1}{2} = 4 = 2

. Therefore, the foci are

S (2, 0)

and

S^{'} (- 2, 0)

The Tangent's Dance

Consider the line

x + 2 y = 0

, which can be rewritten as

y = - \frac{1}{2} x

. Its slope is

- \frac{1}{2}

We seek a tangent perpendicular to this line. Since the product of perpendicular slopes must be

- 1

, our tangent's slope

m

must be

2

Using the condition of tangency

y = m x \pm a^{2} m^{2} + b^{2}

, we substitute

m = 2

a^{2} = 8

, and

b^{2} = 4

y = 2 x \pm 8 (4) + 4 = 2 x \pm 36 = 2 x \pm 6

The problem constrains point

P

to the second quadrant, where

x < 0

and

y > 0

. For a line with a positive slope of

2

to touch the ellipse in this quadrant, it must have a positive

y

-intercept. Thus, we select the tangent

y = 2 x + 6

Pinpointing the Point of Contact

To find the exact coordinates of

P (x_{1}, y_{1})

, we use the point of contact formula for a tangent

y = m x + c

P = (- \frac{a ^{2} m}{c}, \frac{b ^{2}}{c})

Substituting

c = 6

m = 2

a^{2} = 8

, and

b^{2} = 4

x_{1} = - \frac{8 ( 2 )}{6} = - \frac{16}{6} = - \frac{8}{3}

y_{1} = \frac{4}{6} = \frac{2}{3}

Our point of contact is

P (- \frac{8}{3}, \frac{2}{3})

, which correctly lies in the second quadrant.

The Final Area

We calculate the area of triangle

S P S^{'}

. The base

S S^{'}

lies on the

x

-axis between

S (2, 0)

and

S^{'} (- 2, 0)

, giving a base length of

4

The height of the triangle is the perpendicular distance from

P

to the

x

-axis, which is the

y

-coordinate of

P

\frac{2}{3}

. The area

A

is:

A = \frac{1}{2} \times base \times height = \frac{1}{2} \times 4 \times \frac{2}{3} = \frac{4}{3}

Finally, we evaluate the expression

(5 - e^{2}) A

(5 - \frac{1}{2}) \times \frac{4}{3} = \frac{9}{2} \times \frac{4}{3} = \frac{36}{6} = 6

The final result is 6.

Common Traps & FAQs

Wait, why did we choose the positive y-intercept (

y = 2 x + 6

) instead of the negative one (

y = 2 x - 6

Great observation! The problem explicitly places point

P

in the second quadrant. In the second quadrant,

x

is negative and

y

is positive. If we chose the negative intercept, the line would pass through the fourth quadrant (where

y

is negative). By choosing the positive intercept, we ensure the tangent line passes through the upper-left region of the coordinate plane, satisfying the condition.

Is there a faster way to find the point of contact

P

without using the formula

(- \frac{a ^{2} m}{c}, \frac{b ^{2}}{c})

Yes, you could use the parametric form

P (a cos θ, b sin θ)

and equate the slope of the tangent at that point to

m = 2

. However, the formula

(- \frac{a ^{2} m}{c}, \frac{b ^{2}}{c})

is derived directly from the condition of tangency and is often much faster for competitive exams like JEE. It saves precious time and reduces the chance of algebraic errors.

Why is the base of the triangle

S P S^{'}

equal to 4? I thought the distance between foci was

2 a e

You are absolutely correct! The distance between the foci is indeed

2 a e

. Let's check our math:

a^{2} = 8

, so

a = 22

e^{2} = 1 - \frac{b ^{2}}{a ^{2}} = 1 - \frac{4}{8} = \frac{1}{2}

, so

e = \frac{1}{2}

. Therefore,

2 a e = 2 \cdot (22) \cdot (\frac{1}{2}) = 2 \cdot 2 = 4

. The math holds up perfectly!

Select Answer:

Visualized Solution

Ellipse Equation

Eccentricity Formula

Calculating Eccentricity

Locating the Foci

Slope of the Tangent

Equation of Tangent

Simplifying Tangent Equation

Quadrant Constraint

Point of Contact Formula

Calculating Point P

Triangle $Δ S P S^{'}$

Area of Triangle $A$

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent's Dance

Pinpointing the Point of Contact

The Final Area

Similar Questions

If the tangents on the ellipse $4 x^{2} + y^{2} = 8$ at the points $(1, 2)$ and $(a, b)$ are perpendicular to each other, then $a^{2}$ is equal to :

Let the tangents at the points $P$ and $Q$ on the ellipse $\frac{x ^{2}}{2} + \frac{y ^{2}}{4} = 1$ meet at the point $R (2, 22 - 2)$ . If $S$ is the focus of the ellipse on its negative major axis, then $S P^{2} + S Q^{2}$ is equal to

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 :

Let $E$ be an ellipse whose axes are parallel to the co-ordinates axes, having its center at $(3, - 4)$ , one focus at $(4, - 4)$ and one vertex at $(5, - 4)$ . If $m x - y = 4, m > 0$ is a tangent to the ellipse $E$ , then the value of $5 m^{2}$ is equal to

Find the co-ordinates of all the points $P$ on the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ , for which the area of the triangle $P O N$ is a maximum, where $O$ denotes the origin and $N$ , the foot of the perpendicular from $O$ to the tangent at $P$ .

If $3 x + 4 y = 122$ is a tangent to the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{9} = 1$ , for some $a \in R$ then the distance between the foci of the ellipse is :

Select Answer:

Visualized Solution

Ellipse Equation

Eccentricity Formula

Calculating Eccentricity

Locating the Foci

Slope of the Tangent

Equation of Tangent

Simplifying Tangent Equation

Quadrant Constraint

Point of Contact Formula

Calculating Point P

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

Area of Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent's Dance

Pinpointing the Point of Contact

The Final Area

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangents on the ellipse 4x2+y2=8 at the points (1,2) and (a,b) are perpendicular to each other, then a2 is equal to :

.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 :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 3x+4y=122​ is a tangent to the ellipse a2x2​+9y2​=1, for some a∈R then the distance between the foci of the ellipse is :

Triangle $Δ S P S^{'}$

Area of Triangle $A$

If the tangents on the ellipse $4 x^{2} + y^{2} = 8$ at the points $(1, 2)$ and $(a, b)$ are perpendicular to each other, then $a^{2}$ is equal to :

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 :

If $3 x + 4 y = 122$ is a tangent to the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{9} = 1$ , for some $a \in R$ then the distance between the foci of the ellipse is :