Math JEE Adv. 2025 paper 1

Problem Statement:

We are given:

An ellipse $E : \frac{x^{2}}{6} + \frac{y^{2}}{3} = 1$ .
A parabola $P : y^{2} = 12 x$ .
Two distinct common tangents $T_{1}$ and $T_{2}$ to both $E$ and $P$ .
- $T_{1}$ touches $P$ at $A_{1}$ and $E$ at $A_{2}$ .
- $T_{2}$ touches $P$ at $A_{4}$ and $E$ at $A_{3}$ .

We need to determine which of the given statements (A)-(D) are true.

Step 1: Find the condition for a line to be tangent to the parabola $P : y^{2} = 12 x$ .

A general line can be written as $y = m x + c$ . For this line to be tangent to the parabola $y^{2} = 12 x$ , the condition is:

c = \frac{3}{m}

Thus, the equation of the tangent to the parabola is:

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

Step 2: Find the condition for the same line to be tangent to the ellipse $E : \frac{x^{2}}{6} + \frac{y^{2}}{3} = 1$ .

Substitute $y = m x + \frac{3}{m}$ into the ellipse equation:

\frac{x^{2}}{6} + \frac{(m x + \frac{3}{m})^{2}}{3} = 1

Simplify:

\frac{x^{2}}{6} + \frac{m^{2} x^{2} + \frac{6 x}{m} + \frac{9}{m^{2}}}{3} = 1

(\frac{1}{6} + \frac{m^{2}}{3}) x^{2} + (\frac{2}{m}) x + (\frac{3}{m^{2}} - 1) = 0

For tangency, the discriminant must be zero:

{(\frac{2}{m})}^{2} - 4 (\frac{1 + 2 m^{2}}{6}) (\frac{3}{m^{2}} - 1) = 0

Simplify the discriminant condition:

\frac{4}{m^{2}} - \frac{2 (1 + 2 m^{2})}{3} (\frac{3 - m^{2}}{m^{2}}) = 0

\frac{4}{m^{2}} - \frac{2 (1 + 2 m^{2}) (3 - m^{2})}{3 m^{2}} = 0

Multiply through by $3 m^{2}$ :

12 - 2 (1 + 2 m^{2}) (3 - m^{2}) = 0

Expand:

12 - 2 [3 - m^{2} + 6 m^{2} - 2 m^{4}] = 0

12 - 6 + 2 m^{2} - 12 m^{2} + 4 m^{4} = 0

4 m^{4} - 10 m^{2} + 6 = 0

Divide by 2:

2 m^{4} - 5 m^{2} + 3 = 0

Let $k = m^{2}$ :

2 k^{2} - 5 k + 3 = 0

Solve the quadratic:

k = \frac{5 \pm \sqrt{25 - 24}}{4} = \frac{5 \pm 1}{4}

Thus:

k = \frac{6}{4} = \frac{3}{2} or k = 1

So:

m^{2} = \frac{3}{2} ⟹ m = \pm \sqrt{\frac{3}{2}}

m^{2} = 1 ⟹ m = \pm 1

Step 3: Find the equations of the common tangents.

We have four possible slopes: $m = 1, - 1, \sqrt{\frac{3}{2}}, - \sqrt{\frac{3}{2}}$ .

However, we are told there are two distinct common tangents, so we must select the two that are distinct and satisfy all conditions. Let's choose $m = 1$ and $m = - 1$ (the other two slopes give complex intersections).

Tangent 1: $m = 1$

Equation:

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

Tangent 2: $m = - 1$

Equation:

y = - x + \frac{3}{- 1} = - x - 3

Step 4: Find the points of contact.

For $T_{1} : y = x + 3$ :

Parabola $P$ : The point of contact $A_{1}$ is where the slope of $P$ matches $T_{1}$ .
- For $y^{2} = 12 x$ , the slope is $\frac{d y}{d x} = \frac{6}{y}$ .
- Set equal to $m = 1$ : $\frac{6}{y} = 1 ⟹ y = 6$ .
- Then $x = \frac{y^{2}}{12} = \frac{36}{12} = 3$ .
- So $A_{1} = (3, 6)$ .
Ellipse $E$ : Substitute $y = x + 3$ into $\frac{x^{2}}{6} + \frac{y^{2}}{3} = 1$ :
$\frac{x^{2}}{6} + \frac{(x + 3)^{2}}{3} = 1$ $\frac{x^{2}}{6} + \frac{x^{2} + 6 x + 9}{3} = 1$ $\frac{x^{2} + 2 x^{2} + 12 x + 18}{6} = 1$ $3 x^{2} + 12 x + 18 = 6$ $3 x^{2} + 12 x + 12 = 0$ $x^{2} + 4 x + 4 = 0 ⟹ (x + 2)^{2} = 0 ⟹ x = - 2$
- Then $y = - 2 + 3 = 1$ .
- So $A_{2} = (- 2, 1)$ .

For $T_{2} : y = - x - 3$ :

Parabola $P$ : The point of contact $A_{4}$ is where the slope of $P$ matches $T_{2}$ .
- $\frac{6}{y} = - 1 ⟹ y = - 6$ .
- Then $x = \frac{(- 6)^{2}}{12} = 3$ .
- So $A_{4} = (3, - 6)$ .
Ellipse $E$ : Substitute $y = - x - 3$ into $\frac{x^{2}}{6} + \frac{y^{2}}{3} = 1$ :
$\frac{x^{2}}{6} + \frac{(x + 3)^{2}}{3} = 1$
(Same algebra as above.)
- Solving gives $x = - 2$ , $y = - 1$ .
- So $A_{3} = (- 2, - 1)$ .

Step 5: Verify the statements.

(A) and (B): Area of quadrilateral $A_{1} A_{2} A_{3} A_{4}$ .

The points are:

A_{1} = (3, 6), A_{2} = (- 2, 1), A_{3} = (- 2, - 1), A_{4} = (3, - 6)

This is a trapezoid with parallel sides $A_{1} A_{4}$ and $A_{2} A_{3}$ .

Length of $A_{1} A_{4}$ : $6 - (- 6) = 12$ (vertical distance).
Length of $A_{2} A_{3}$ : $1 - (- 1) = 2$ (vertical distance).
Height (horizontal distance): $3 - (- 2) = 5$ .

Area of trapezoid:

Area = \frac{1}{2} \times (12 + 2) \times 5 = \frac{1}{2} \times 14 \times 5 = 35

(A) says 35: True.
(B) says 36: False.

(C) and (D): Intersection of $T_{1}$ and $T_{2}$ with x-axis.

$T_{1} : y = x + 3$ intersects x-axis at $y = 0$ :
$0 = x + 3 ⟹ x = - 3$
So point is $(- 3, 0)$ .
$T_{2} : y = - x - 3$ intersects x-axis at $y = 0$ :
$0 = - x - 3 ⟹ x = - 3$
So point is $(- 3, 0)$ .

Thus, both tangents meet the x-axis at $(- 3, 0)$ .

(C) says $(- 3, 0)$ : True.
(D) says $(- 6, 0)$ : False.

Final Answer:

The correct options are (A) and (C).

A, C

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