Question

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

• A. \( \tan^{-1} \left(\frac{11}{9} \right) \)
• B. \( \frac{\pi}{2} - \tan^{-1} \left(\frac{3}{2} \right) \)
• C. \( \tan^{-1} \left(\frac{4}{5} \right) \)
• D. \( \tan^{-1} \left(\frac{9}{7} \right) \)

Hint:

For angles subtended by chords at the vertex of a parabola, use the tangent formula to determine the required angle.

Answer 1

The Correct Option is D

To find the angle that the line segment \( AB \) subtends at the vertex of the parabola, we first need to determine the points of intersection \( A \) and \( B \) where the line and the parabola intersect. 

1. Rewrite the equation of the line and the parabola for ease of calculation:
   \(3x - 2y + 12 = 0 \Longrightarrow y = \frac{3}{2}x + 6\)
   The equation of the parabola is \(4y = 3x^2 \Longrightarrow y = \frac{3}{4}x^2\).
2. Set the two expressions for \( y \) equal to find the \( x \)-coordinates of points \( A \) and \( B \):
   \(\frac{3}{4}x^2 = \frac{3}{2}x + 6\)
   Multiplying every term by 4 to eliminate fractions: 
   \(3x^2 = 6x + 24\)
   Rearranging gives: \(3x^2 - 6x - 24 = 0\).
3. Simplify and solve the quadratic equation:
   Divide the entire equation by 3: \(x^2 - 2x - 8 = 0\).
   Factorize: \((x - 4)(x + 2) = 0\).
   The solutions are \( x = 4 \) and \( x = -2 \).
4. Substitute back to find \( y \)-coordinates:
   For \( x = 4 \), \(y = \frac{3}{4}(4)^2 = 12\).
   For \( x = -2 \), \(y = \frac{3}{4}(-2)^2 = 3\).
5. Hence, the points A and B are \( A(4, 12) \) and \( B(-2, 3) \).
6. Determine the coordinates of the vertex of the parabola. For \( y = \frac{3}{4}x^2 \), the vertex is at the origin, \( V(0, 0) \).
7. Calculate the slopes of the lines \( VA \) and \( VB \):
   • Slope of \( VA \): \(m_1 = \frac{12 - 0}{4 - 0} = 3\).
   • Slope of \( VB \): \(m_2 = \frac{3 - 0}{-2 - 0} = -\frac{3}{2}\).
8. Calculate the angle \( \theta \) between \( VA \) and \( VB \) using the formula for angle between two lines: \(\theta = \tan^{-1}\left( \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \right)\) 
   Substituting the values, we get: \(\theta = \tan^{-1}\left( \left| \frac{3 - \left(-\frac{3}{2}\right)}{1 + 3 \left(-\frac{3}{2}\right)} \right| \right)\) 
   Simplifying: \(\theta = \tan^{-1}\left( \frac{\frac{9}{2}}{-\frac{7}{2}} \right) = \tan^{-1}\left( \frac{9}{7} \right)\)

Therefore, the correct answer is \(\tan^{-1} \left(\frac{9}{7} \right)\).

Answer 2

To solve the problem, we need to determine the intersection points of the line \(3x - 2y + 12 = 0\) with the parabola \(4y = 3x^2\), and then find the angle subtended by the line segment connecting these points at the vertex of the parabola.

The parabola is given by \(4y = 3x^2\) or \(y = \frac{3}{4}x^2\). 
We substitute this into the line's equation:

\(3x - 2\left(\frac{3}{4}x^2\right) + 12 = 0\)

Simplifying gives:

\(3x - \frac{3}{2}x^2 + 12 = 0\)

Multiplying the entire equation by 2 to eliminate fractions:

\(6x - 3x^2 + 24 = 0\)

Rearrange to a standard quadratic form:

\(3x^2 - 6x - 24 = 0\)

Divide the equation by 3:

\(x^2 - 2x - 8 = 0\)

Factor the quadratic equation:

\((x - 4)(x + 2) = 0\)

Hence, the solutions for \(x\) are \(x = 4\) and \(x = -2\).

Substitute these \(x\) values back into the parabola's equation \(y = \frac{3}{4}x^2\) to find corresponding \(y\) values:

For \(x = 4\):

\(y = \frac{3}{4}(4)^2 = 12\)

For \(x = -2\):

\(y = \frac{3}{4}(-2)^2 = 3\)

Thus, the points of intersection are \(A(4, 12)\) and \(B(-2, 3)\).

The vertex of the parabola \(4y = 3x^2\) is at \((0, 0)\).

We want to find the angle subtended by line segment \(AB\) at the vertex. The slope of line \(AB\) is given by:

\(m = \frac{12 - 3}{4 - (-2)} = \frac{9}{6} = \frac{3}{2}\)

The angle \(\theta\) that line \(AB\) makes with the horizontal is:

\(\theta = \tan^{-1}\left(\frac{3}{2}\right)\)

The angle subtended at the origin is defined by the angle between the lines \(y = \frac{3}{2}x\) and the negative of this line representing direction opposite to the vector \((x, y) = (4, 12)\).

This gives an angle of:

\(\theta = 2\tan^{-1}\left(\frac{3}{2}\right)\)

From trigonometric identities, we know the angle between the x-axis and the line segment is:

\(\phi = \pi - 2\theta = \pi - 2\tan^{-1}\left(\frac{3}{2}\right)\)

By identity, the angle subtended is half of \(\phi\):

\(\theta = \tan^{-1}\left(\frac{2m}{1-m^2}\right)\), simplifying for \(m = \frac{3}{2}\) gives the subtended angle.

\(\frac{2\times\frac{3}{2}}{1-\left(\frac{3}{2}\right)^2} = \tan^{-1}\left(\frac{9}{7}\right)\)

Thus, the line segment \(AB\) subtends an angle of \(\tan^{-1} \left(\frac{9}{7}\right)\) at the vertex of the parabola.