If the line 3x – 2y + 12 = 0 intersects the parabola at the points A and B, then at the vertex of the parabola, the line segment AB subtends and angle equal to [2025]

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Q.
If the line 3x – 2y + 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 [2025]

1 $\tan^{- 1} (\frac{11}{9})$

2 $\tan^{- 1} (\frac{9}{7})$

3 $\tan^{- 1} (\frac{4}{5})$

4 $\frac{π}{2} - \tan^{- 1} (\frac{3}{2})$

Ans.
(2)

Given, 3x – 2y + 12 = 0

$\Rightarrow y = \frac{3 x + 12}{2}$

Put value of 'y' in $4 y = 3 x^{2}$ , we get

$\Rightarrow 2 (3 x + 12) = 3 x^{2}$

$\Rightarrow x^{2} - 2 x - 8 = 0 \Rightarrow x = - 2, 4$

$∴$ y = 3, 12

Let A(–2, 3) and B(4, 12)

Since, vertex of parabola is O(0, 0).

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

$∴ \tan θ =$ $| \frac{m_{O A} - m_{O B}}{1 + m_{O A} \times m_{O B}} |$ $=$ $| \frac{\frac{- 3}{2} - 3}{1 + (\frac{- 3}{2}) \times 3} |$

$\Rightarrow \tan θ = \frac{9}{7} \Rightarrow θ = \tan^{- 1} (\frac{9}{7})$ .

Solution

Q.
If the line 3x – 2y + 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 [2025]

1 $\tan^{- 1} (\frac{11}{9})$

2 $\tan^{- 1} (\frac{9}{7})$

3 $\tan^{- 1} (\frac{4}{5})$

4 $\frac{π}{2} - \tan^{- 1} (\frac{3}{2})$

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Solution

Q. If the line 3x – 2y + 12 = 0 intersects the parabola 4y=3x2 at the points A and B, then at the vertex of the parabola, the line segment AB subtends an angle equal to [2025]

1 tan–1(119)

2 tan–1(97)

3 tan–1(45)

4 π2–tan–1(32)

Want Us to Email you About Special Offers & Updates?

Q.
If the line 3x – 2y + 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 [2025]

1 $\tan^{- 1} (\frac{11}{9})$

2 $\tan^{- 1} (\frac{9}{7})$

3 $\tan^{- 1} (\frac{4}{5})$

4 $\frac{π}{2} - \tan^{- 1} (\frac{3}{2})$