Simplify the equation to get a quadratic equation in terms of x. After simplifying, we get 3x² - 6x - 24 = 0, which can be further simplified to x² - 2x - 8 = 0. Factoring this quadratic equation gives us (x - 4)(x + 2) = 0, yielding x -values of 4 and -2.

Simplify the equation to get a quadratic equation in terms of x. After simplifying, we get 3x² - 6x - 24 = 0, which can be further simplified to x² - 2x - 8 = 0. Factoring this quadratic equation gives us (x - 4)(x + 2) = 0, yielding x -values of 4 and -2.

Now, substitute these x -values back into the linear equation y = (3x + 12)/(2) to find the corresponding y -values. For x=4, y=12, so point A is (4, 12). For x=-2, y=3, so point B is (-2, 3).

The equation of the parabola is 4y = 3x², which can be written as y = (3)/(4)x². This is a parabola of the form y = ax², which has its vertex at the origin (0, 0). Let's call the vertex V(0, 0).

The angle θ between two lines with slopes m_1 and m_2 is given by the formula tan θ = |(m_1 - m_2)/(1 + m_1 m_2)|. Substituting the slopes m_VA = 3 and m_VB = -(3)/(2), we get tan θ = (9)/(7). Therefore, the angle θ is tan^(-1)((9)/(7)).

The angle θ between two lines with slopes m_1 and m_2 is given by the formula tan θ = |(m_1 - m_2)/(1 + m_1 m_2)|. Substituting the slopes m_VA = 3 and m_VB = -(3)/(2), we get tan θ = (9)/(7). Therefore, the angle θ is tan^(-1)((9)/(7)).

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

Answer: The angle subtended by the line segment $AB$ at the vertex of the parabola is $\tan^{-1}\left(\frac{9}{7}\right)$ .

Step-by-step solution

Step 1: Find the intersection points A and B

First, we need to find the coordinates of the intersection points A and B. We can do this by substituting the expression for $y$ from the linear equation into the parabolic equation. The line equation is $3x - 2y + 12 = 0$ , which can be rewritten as $y = \frac{3x + 12}{2}$ . Substitute this into the parabola equation $4y = 3x^2$ .

Step 2: Solve the quadratic equation for x

Simplify the equation to get a quadratic equation in terms of $x$ . After simplifying, we get $3x^2 - 6x - 24 = 0$ , which can be further simplified to $x^2 - 2x - 8 = 0$ . Factoring this quadratic equation gives us $(x - 4)(x + 2) = 0$ , yielding $x$ -values of $4$ and $-2$ .

Step 3: Find the coordinates of A and B

Now, substitute these $x$ -values back into the linear equation $y = \frac{3x + 12}{2}$ to find the corresponding $y$ -values. For $x=4$ , $y=12$ , so point A is $(4, 12)$ . For $x=-2$ , $y=3$ , so point B is $(-2, 3)$ .

Step 4: Identify the vertex of the parabola

The equation of the parabola is $4y = 3x^2$ , which can be written as $y = \frac{3}{4}x^2$ . This is a parabola of the form $y = ax^2$ , which has its vertex at the origin $(0, 0)$ . Let's call the vertex $V(0, 0)$ .

Step 5: Calculate the slopes of VA and VB

To find the angle subtended by the line segment AB at the vertex V, we need to find the slopes of the lines VA and VB. The slope of VA is $\frac{12-0}{4-0} = 3$ . The slope of VB is $\frac{3-0}{-2-0} = -\frac{3}{2}$ .

Step 6: Calculate the angle using the tangent formula

The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by the formula $\tan \theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$ . Substituting the slopes $m_{VA} = 3$ and $m_{VB} = -\frac{3}{2}$ , we get $\tan \theta = \frac{9}{7}$ . Therefore, the angle $\theta$ is $\tan^{-1}\left(\frac{9}{7}\right)$ .

More from this chapter

All Conic Sections solutions →

Solve your own maths question free →

If the line 3x−2y+12=03x - 2y + 12 = 03x−2y+12=0 intersects the parabola 4y=3x24y = 3x^24y=3x2 at the points AAA and BBB, then at the vertex of the parabola, the line segment ABABAB subtends an angle equal to