JEE Advanced 2023

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: Let $T_{1}$ and $T_{2}$ be two distinct common tangents to the ellipse $E : \frac{x ^{2}}{6} + \frac{y ^{2}}{3} = 1$ and the parabola $P : y^{2} = 12 x$ . Suppose that the tangent $T_{1}$ touches $P$ and $E$ at the point $A_{1}$ and $A_{2}$ , respectively and the tangent $T_{2}$ touches $P$ and $E$ at the points $A_{4}$ and $A_{3}$ , respectively. Then which of the following statements is(are) true?

Q: Why do we use $c = a/m$ for the parabola? Can't we use calculus?

You absolutely can! Using $dy/dx$ is a valid path. However, in JEE Advanced, time is your currency. Memorizing the standard tangent form $y = mx + a/m$ saves you precious minutes and reduces the chance of calculation errors during differentiation. It is a powerful tool to keep in your arsenal.

Q: Why is the quadrilateral a trapezium?

Look at the coordinates of the points of contact. $A_1$ and $A_4$ share the same x-coordinate, as do $A_2$ and $A_3$. This means the sides $A_1A_4$ and $A_2A_3$ are vertical lines, which are parallel to the y-axis. Since they are parallel to each other, the shape is a trapezium! This geometric insight turns a complex area problem into a simple calculation.

Q: Why do we reject the negative root in the biquadratic?

Actually, we don't! We found $m^2 = 1$, which gives $m = 1$ and $m = -1$. Both are valid slopes. The negative slope gives us the second tangent, which is just the reflection of the first across the x-axis. Both are necessary to form the quadrilateral, so we keep both!

Select Answer:

* Multiple Correct

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the geometry. You have a parabola opening to the right and an ellipse centered at the origin. A common tangent is a line that kisses both curves simultaneously. Before diving into the algebra, sketch the curves to see where these lines might lie. Notice the symmetry—the curves are symmetric about the x-axis, so expect the tangents to be symmetric as well.

Hint 2

Hint 2 (The Arsenal): Recall your toolkit for tangency. For a parabola

y^{2} = 4 a x

, the tangent is

y = m x + \frac{a}{m}

. For an ellipse

\frac{x ^{2}}{A ^{2}} + \frac{y ^{2}}{B ^{2}} = 1

, the condition for tangency is

c^{2} = A^{2} m^{2} + B^{2}

. You need to equate the

c

values from both conditions to find the slope

m

Hint 3

Hint 3 (The Execution): Start by setting the

c

from the parabola tangent equal to the

c

required by the ellipse. You will get

\frac{3}{m} = \pm 6 m^{2} + 3

. Square both sides to eliminate the radical and solve the resulting biquadratic equation for

m

. Once you have

m

, finding the points of contact is just a matter of plugging

m

into the standard coordinate formulas.

Visualized Solution

Visual Anchor: $P$ and $E$

Given curves:
* Parabola $P : y^{2} = 12 x$
* Ellipse $E : \frac{x ^{2}}{6} + \frac{y ^{2}}{3} = 1$

Tangent to Parabola $P$

Tangent to parabola $y^{2} = 4 a x$ with slope $m$ :
$y = m x + \frac{a}{m}$
Here, $4 a = 12 ⟹ a = 3$
Equation: $y = m x + \frac{3}{m}$

Tangency Condition for Ellipse $E$

Condition for tangency to ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ :
$c^{2} = a^{2} m^{2} + b^{2}$
Substitute $c = \frac{3}{m}$ , $a^{2} = 6$ , $b^{2} = 3$ :
$(\frac{3}{m})^{2} = 6 m^{2} + 3$

Simplifying the Equation

$\frac{9}{m ^{2}} = 6 m^{2} + 3$
Multiply by $m^{2}$ :
$9 = 6 m^{4} + 3 m^{2}$
Rearrange terms:
$6 m^{4} + 3 m^{2} - 9 = 0$

Solving for Slope $m$

Divide by $3$ :
$2 m^{4} + m^{2} - 3 = 0$
Factorize:
$(2 m^{2} + 3) (m^{2} - 1) = 0$
Since $m \in R$ , $2 m^{2} + 3 \neq = 0$
$⟹ m^{2} = 1 ⟹ m = \pm 1$

Equations of Tangents $T_{1}, T_{2}$

Substitute $m = \pm 1$ into $y = m x + \frac{3}{m}$ :
For $m = 1$ : $T_{1} : y = x + 3$
For $m = - 1$ : $T_{2} : y = - x - 3$

Intersection with X-axis

Intersection with x-axis ( $y = 0$ ):
$0 = x + 3 ⟹ x = - 3$
$0 = - x - 3 ⟹ x = - 3$
Point of intersection: $(- 3, 0)$

Point of Contact Formula for $P$

Point of contact on parabola $y^{2} = 4 a x$ :
$P (m) = (\frac{a}{m ^{2}}, \frac{2 a}{m})$

Evaluating $A_{1}$ and $A_{4}$

Substitute $a = 3$ and $m = \pm 1$ :
For $m = 1$ : $A_{1} = (\frac{3}{1 ^{2}}, \frac{6}{1}) = (3, 6)$
For $m = - 1$ : $A_{4} = (\frac{3}{( - 1 ) ^{2}}, \frac{6}{- 1}) = (3, - 6)$

Point of Contact Formula for $E$

Point of contact on ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ :
$E (m) = (- \frac{a ^{2} m}{c}, \frac{b ^{2}}{c})$
Recall $c = \frac{3}{m}$

Evaluating $A_{2}$ and $A_{3}$

For $m = 1, c = 3$ :
$A_{2} = (- \frac{6 ( 1 )}{3}, \frac{3}{3}) = (- 2, 1)$
For $m = - 1, c = - 3$ :
$A_{3} = (- \frac{6 ( - 1 )}{- 3}, \frac{3}{- 3}) = (- 2, - 1)$

Forming Quadrilateral $A_{1} A_{2} A_{3} A_{4}$

Quadrilateral $A_{1} A_{2} A_{3} A_{4}$ :
$A_{1} (3, 6), A_{2} (- 2, 1), A_{3} (- 2, - 1), A_{4} (3, - 6)$
Notice $x$ -coordinates:
$A_{1}, A_{4}$ have $x = 3$
$A_{2}, A_{3}$ have $x = - 2$

Identifying the Shape

Shape: Trapezium (since $A_{1} A_{4} ∥ A_{2} A_{3}$ )
$Area = \frac{1}{2} \times (Sum of parallel sides) \times height$

Calculating Dimensions

Parallel sides:
$A_{1} A_{4} = 6 - (- 6) = 12$
$A_{2} A_{3} = 1 - (- 1) = 2$
Height (distance between $x = 3$ and $x = - 2$ ):
$h = 3 - (- 2) = 5$

Final Area Calculation

$Area = \frac{1}{2} \times (12 + 2) \times 5$
$Area = \frac{1}{2} \times 14 \times 5 = 35 sq. units$
Key Takeaways:
* Common tangents require equating tangency conditions.
* Geometric visualization simplifies area calculation.
Correct Options: A and C*.

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

We are tasked with finding the common tangents to the parabola

P : y^{2} = 12 x

and the ellipse

E : \frac{x ^{2}}{6} + \frac{y ^{2}}{3} = 1

For the parabola

y^{2} = 12 x

, we identify

4 a = 12

, which implies

a = 3

. The equation of any tangent to this parabola with slope

m

is given by:

y = m x + \frac{3}{m}

The Master Equation

For a line

y = m x + c

to be tangent to the ellipse

\frac{x ^{2}}{6} + \frac{y ^{2}}{3} = 1

, it must satisfy the condition

c^{2} = A^{2} m^{2} + B^{2}

. Here,

A^{2} = 6

and

B^{2} = 3

Substituting

c = \frac{3}{m}

into the tangency condition

c^{2} = 6 m^{2} + 3

, we obtain:

(\frac{3}{m})^{2} = 6 m^{2} + 3

Simplifying this expression leads to:

\frac{9}{m ^{2}} = 6 m^{2} + 3 \Rightarrow 9 = 6 m^{4} + 3 m^{2}

Dividing by 3, we arrive at the biquadratic equation:

2 m^{4} + m^{2} - 3 = 0

Factoring the quadratic in terms of

m^{2}

, we get

(2 m^{2} + 3) (m^{2} - 1) = 0

. Since

m

must be real, $2m^2 + 3 eq 0$, which leaves us with

m^{2} = 1

. Thus, the slopes are $m = 1$ and $m = - 1$ .

Defining the Tangents

Using the slopes

m = 1

and

m = - 1

, we define the two common tangents:

For

m = 1

, the tangent is

T_{1} : y = x + 3

For

m = - 1

, the tangent is

T_{2} : y = - x - 3

Setting

y = 0

in both equations, we find that both tangents intersect the x-axis at the point $(- 3, 0)$ .

Points of Contact

For the parabola, the point of contact is given by

(\frac{a}{m ^{2}}, \frac{2 a}{m})

. Substituting

a = 3

and

m = \pm 1

, we find the points:

A_{1} (3, 6) and A_{4} (3, - 6)

For the ellipse, the point of contact is given by

(- \frac{A ^{2} m}{c}, \frac{B ^{2}}{c})

. Substituting

A^{2} = 6, B^{2} = 3, c = \pm 3

, we find the points:

A_{2} (- 2, 1) and A_{3} (- 2, - 1)

Final Calculation

The points

A_{1} (3, 6)

A_{2} (- 2, 1)

A_{3} (- 2, - 1)

, and

A_{4} (3, - 6)

form a trapezium. The parallel sides are vertical segments

A_{1} A_{4}

and

A_{2} A_{3}

The length of

A_{1} A_{4}

∣6 - (- 6) ∣ = 12

. The length of

A_{2} A_{3}

∣1 - (- 1) ∣ = 2

The height of the trapezium is the horizontal distance between

x = 3

and

x = - 2

, which is

h = 3 - (- 2) = 5

. The area is calculated as:

Area = \frac{1}{2} \times (12 + 2) \times 5 = 35

The final area of the quadrilateral formed by the points of contact is $35$ square units.

Common Traps & FAQs

Why do we use

c = a / m

for the parabola? Can't we use calculus?

You absolutely can! Using

d y / d x

is a valid path. However, in JEE Advanced, time is your currency. Memorizing the standard tangent form

y = m x + a / m

saves you precious minutes and reduces the chance of calculation errors during differentiation. It is a powerful tool to keep in your arsenal.

Why is the quadrilateral a trapezium?

Look at the coordinates of the points of contact.

A_{1}

and

A_{4}

share the same x-coordinate, as do

A_{2}

and

A_{3}

. This means the sides

A_{1} A_{4}

and

A_{2} A_{3}

are vertical lines, which are parallel to the y-axis. Since they are parallel to each other, the shape is a trapezium! This geometric insight turns a complex area problem into a simple calculation.

Why do we reject the negative root in the biquadratic?

Actually, we don't! We found

m^{2} = 1

, which gives

m = 1

and

m = - 1

. Both are valid slopes. The negative slope gives us the second tangent, which is just the reflection of the first across the x-axis. Both are necessary to form the quadrilateral, so we keep both!

Select Answer:

Visualized Solution

Visual Anchor: $P$ and $E$

Tangent to Parabola $P$

Tangency Condition for Ellipse $E$

Simplifying the Equation

Solving for Slope $m$

Equations of Tangents $T_{1}, T_{2}$

Intersection with X-axis

Point of Contact Formula for $P$

Evaluating $A_{1}$ and $A_{4}$

Point of Contact Formula for $E$

Evaluating $A_{2}$ and $A_{3}$

Forming Quadrilateral $A_{1} A_{2} A_{3} A_{4}$

Identifying the Shape

Calculating Dimensions

Final Area Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Master Equation

Defining the Tangents

Points of Contact

Final Calculation

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Select Answer:

Visualized Solution

Visual Anchor: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E

Tangent to Parabola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Tangency Condition for Ellipse .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E

Simplifying the Equation

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

Equations of Tangents .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T1​,T2​

Intersection with X-axis

Point of Contact Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A4​

Point of Contact Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A3​

Forming Quadrilateral .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​A2​A3​A4​

Identifying the Shape

Calculating Dimensions

Final Area Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Master Equation

Defining the Tangents

Points of Contact

Final Calculation

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Tangent to Parabola $P$

Tangency Condition for Ellipse $E$

Solving for Slope $m$

Equations of Tangents $T_{1}, T_{2}$

Point of Contact Formula for $P$

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