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.
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