JEE Main 2025 Jan 23 Shift 1, Mathematics Q4: Parabola: Chord, Tangent and Normal

JEE Main2025Jan 23, Shift 1Mathematics

If the line $3 x - 2 y + 12 = 0$ intersects the parabola $4 y = 3 x^{2}$ at the points A and B, then at the vertex of the parabola, the line segment AB subtends an angle equal to

A
$tan^{- 1} (\frac{11}{9})$
B
$\frac{π}{2} - tan^{- 1} (\frac{3}{2})$
C
$tan^{- 1} (\frac{4}{5})$
D
$tan^{- 1} (\frac{9}{7})$

Solution

From the line, $2 y = 3 x + 12$ . Substituting into $4 y = 3 x^{2}$ gives $2 (3 x + 12) = 3 x^{2}$ , so $3 x^{2} - 6 x - 24 = 0 \Rightarrow x^{2} - 2 x - 8 = 0$ , yielding $x = - 2, 4$ .

Corresponding $y$ values: at $x = - 2$ , $y = 3$ ; at $x = 4$ , $y = 12$ . So $A = (- 2, 3)$ and $B = (4, 12)$ .

The vertex of the parabola is the origin. Slopes from $O$ :

$m_{O A} = \frac{3}{- 2} = - \frac{3}{2}$ , $m_{O B} = \frac{12}{4} = 3$ .

$tan θ = \frac{m _{O A} - m _{O B}}{1 + m _{O A} m _{O B}} = \frac{- 3/2 - 3}{1 - 9/2} = \frac{- 9/2}{- 7/2} = \frac{9}{7}$ .

$∴ θ = tan^{- 1} \frac{9}{7}$ (the acute angle).

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JEE Main 2025 Jan 23 Shift 1, Mathematics Q4: Parabola: Chord, Tangent and Normal

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