Maths Q10 – JEE-MAIN 2025

Let for two distinct values of p the lines

y = x + p

touch the ellipse E:

\frac{x^{2}}{4^{2}} + \frac{y^{2}}{3^{2}} = 1

at the points A and B. Let the line

y = x

intersect E at the points C and D. Then the area of the quadrilateral ABCD is equal to :

Clue 1 of 3: Tangency Condition

Recall the condition for a line $y = m x + c$ to be tangent to an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ .

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Ninja Strategy•Maximum Inscribed Area

Recognize that the maximum area of a quadrilateral inscribed in an ellipse is $2 a b$ , which immediately eliminates options greater than 24.

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Step 1: Find Tangent Lines and Points A, B✦ Active

The ellipse is E: $\frac{x^{2}}{4^{2}} + \frac{y^{2}}{3^{2}} = 1$ , so $a^{2} = 16$ and $b^{2} = 9$ . The line is $y = x + p$ , so $m = 1$ and $c = p$ . The tangency condition for an ellipse is $c^{2} = a^{2} m^{2} + b^{2}$ . Substituting the values:

p^{2} = (16) (1)^{2} + 9 = 25 ⟹ p = \pm 5

The tangent points $(x_{0}, y_{0})$ are given by $x_{0} = - \frac{a^{2} m}{c}$ and $y_{0} = \frac{b^{2}}{c}$ . For $p = 5$ , point A is $(- \frac{16 \cdot 1}{5}, \frac{9}{5}) = (- \frac{16}{5}, \frac{9}{5})$ . For $p = - 5$ , point B is $(- \frac{16 \cdot 1}{- 5}, \frac{9}{- 5}) = (\frac{16}{5}, - \frac{9}{5})$ .

Step 2: Find Intersection Points C, D○ Expand

Substitute $y = x$ into the ellipse equation:

\frac{x^{2}}{16} + \frac{x^{2}}{9} = 1 ⟹ 9 x^{2} + 16 x^{2} = 144 ⟹ 25 x^{2} = 144 ⟹ x = \pm \frac{12}{5}

Since $y = x$ , the points are $C = (\frac{12}{5}, \frac{12}{5})$ and $D = (- \frac{12}{5}, - \frac{12}{5})$ .

Step 3: Calculate Area of Quadrilateral ABCD○ Expand

The quadrilateral ABCD can be divided into two triangles, $△ A C D$ and $△ B C D$ , sharing the common base CD. The line segment CD lies on the line $y = x$ (or $x - y = 0$ ). Length of base $C D = \sqrt{(\frac{12}{5} - (- \frac{12}{5}))^{2} + (\frac{12}{5} - (- \frac{12}{5}))^{2}} = \sqrt{(\frac{24}{5})^{2} + (\frac{24}{5})^{2}} = \frac{24 \sqrt{2}}{5}$ . Height from A to $y = x$ ( $h_{A}$ ) = $\frac{| - \frac{16}{5} - \frac{9}{5} |}{\sqrt{1^{2} + (- 1)^{2}}} = \frac{| - 5 |}{\sqrt{2}} = \frac{5}{\sqrt{2}}$ . Height from B to $y = x$ ( $h_{B}$ ) = $\frac{| \frac{16}{5} - (- \frac{9}{5}) |}{\sqrt{1^{2} + (- 1)^{2}}} = \frac{| 5 |}{\sqrt{2}} = \frac{5}{\sqrt{2}}$ . Area of ABCD = $\frac{1}{2} \cdot C D \cdot (h_{A} + h_{B})$ .

Area = \frac{1}{2} \cdot \frac{24 \sqrt{2}}{5} \cdot (\frac{5}{\sqrt{2}} + \frac{5}{\sqrt{2}}) = \frac{1}{2} \cdot \frac{24 \sqrt{2}}{5} \cdot \frac{10}{\sqrt{2}} = \frac{1}{2} \cdot \frac{24}{5} \cdot 10 = 24

💡 Teacher's Secret Hint

Notice that the line $y = x$ is perpendicular to the line connecting the tangent points of $y = x + p$ and $y = x - p$ if the ellipse is a circle. Here, the line $y = x$ is parallel to the tangent lines $y = x + 5$ and $y = x - 5$ .

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Maths Question 10 – JEE-MAIN 2025

Clue 1 of 3: Tangency Condition

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