Question

A parabola having its axis parallel to the Y-axis passes through the points \(\left(0, \frac{2}{5}\right)\), \((4, -2)\), and \(\left(1, \frac{8}{5}\right)\). Then the point that lies on this parabola is:

• A. \(\left(3, \frac{5}{2}\right)\)
• B. \((-1, 2)\)
• C. \(\left(-2, \frac{28}{5}\right)\)
• D. \(\left(2, \frac{8}{5}\right)\)

Hint:

For a parabola with axis parallel to the Y-axis, use the form \( y = ax^2 + bx + c \). Substitute given points to find the coefficients, then test options by substituting them into the equation.

Answer 1

The Correct Option is D

Step 1: Write the general equation of the parabola.
Since the axis is parallel to the Y-axis, the parabola has the form: \[ y = ax^2 + bx + c \] We need to find \(a\), \(b\), and \(c\) using the given points.
Step 2: Substitute the points to form equations.
For \(\left(0, \frac{2}{5}\right)\): \[ \frac{2}{5} = a(0)^2 + b(0) + c \implies c = \frac{2}{5} \quad (1) \] For \((4, -2)\): \[ -2 = a(4)^2 + b(4) + c \implies -2 = 16a + 4b + c \quad (2) \] For \(\left(1, \frac{8}{5}\right)\): \[ \frac{8}{5} = a(1)^2 + b(1) + c \implies \frac{8}{5} = a + b + c \quad (3) \]
Step 3: Solve the system of equations.
From (1), \(c = \frac{2}{5}\). Substitute into (2) and (3):
Equation (2): \[ -2 = 16a + 4b + \frac{2}{5} \implies 16a + 4b = -2 - \frac{2}{5} = -\frac{12}{5} \implies 4a + b = -\frac{3}{5} \quad (4) \] Equation (3): \[ \frac{8}{5} = a + b + \frac{2}{5} \implies a + b = \frac{8}{5} - \frac{2}{5} = \frac{6}{5} \quad (5) \] Subtract (4) from (5): \[ (a + b) - (4a + b) = \frac{6}{5} - \left(-\frac{3}{5}\right) \implies -3a = \frac{9}{5} \implies a = -\frac{3}{5} \] Substitute \(a = -\frac{3}{5}\) into (5): \[ -\frac{3}{5} + b = \frac{6}{5} \implies b = \frac{6}{5} + \frac{3}{5} = \frac{9}{5} \] So, \(a = -\frac{3}{5}\), \(b = \frac{9}{5}\), \(c = \frac{2}{5}\). The equation of the parabola is: \[ y = -\frac{3}{5}x^2 + \frac{9}{5}x + \frac{2}{5} \]
Step 4: Test the options to find which point lies on the parabola.
Option (1) \(\left(3, \frac{5}{2}\right)\): \[ y = -\frac{3}{5}(3)^2 + \frac{9}{5}(3) + \frac{2}{5} = -\frac{27}{5} + \frac{27}{5} + \frac{2}{5} = \frac{2}{5} \neq \frac{5}{2} \] Option (2) \((-1, 2)\): \[ y = -\frac{3}{5}(-1)^2 + \frac{9}{5}(-1) + \frac{2}{5} = -\frac{3}{5} - \frac{9}{5} + \frac{2}{5} = -\frac{10}{5} = -2 \neq 2 \] Option (3) \(\left(-2, \frac{28}{5}\right)\): \[ y = -\frac{3}{5}(-2)^2 + \frac{9}{5}(-2) + \frac{2}{5} = -\frac{12}{5} - \frac{18}{5} + \frac{2}{5} = -\frac{28}{5} \neq \frac{28}{5} \] Option (4) \(\left(2, \frac{8}{5}\right)\): \[ y = -\frac{3}{5}(2)^2 + \frac{9}{5}(2) + \frac{2}{5} = -\frac{12}{5} + \frac{18}{5} + \frac{2}{5} = \frac{8}{5} \] This matches, so \((2, \frac{8}{5})\) lies on the parabola. Final Answer: \[ \boxed{4} \]