Question

If a parabola having its axis parallel to X-axis passes through the points \( (0, -1), (6, 1) \) and \( (-2, -3) \), then the point at which this parabola cuts the X-axis is

• A. \( \left( \frac{5}{2}, 0 \right) \)
• B. \( (-1, 0) \)
• C. \( (6, 0) \)
• D. \( \left( \frac{8}{5}, 0 \right) \)

Hint:

For a parabola with its axis parallel to the X-axis, the equation is of the form \( x = Ay^2 + By + C \). Substitute the coordinates of the given points into this equation to form a system of linear equations in \( A, B, \) and \( C \). Solve this system to find the values of \( A, B, \) and \( C \). To find the point where the parabola cuts the X-axis, set \( y = 0 \) in the equation of the parabola and solve for \( x \).

Answer 1

The Correct Option is A

The equation of a parabola with its axis parallel to the X-axis is of the form \( y = a(x - h)^2 + k \) or \( x = Ay^2 + By + C \).
Since three points are given, the second form is more convenient.
Let the equation of the parabola be \( x = Ay^2 + By + C \).
Since the parabola passes through \( (0, -1) \): \( 0 = A(-1)^2 + B(-1) + C \implies A - B + C = 0 \) (1) Since the parabola passes through \( (6, 1) \): \( 6 = A(1)^2 + B(1) + C \implies A + B + C = 6 \) (2) Since the parabola passes through \( (-2, -3) \): \( -2 = A(-3)^2 + B(-3) + C \implies 9A - 3B + C = -2 \) (3) Subtract (1) from (2): \( (A + B + C) - (A - B + C) = 6 - 0 \implies 2B = 6 \implies B = 3 \) Substitute \( B = 3 \) into (1) and (2): \( A - 3 + C = 0 \implies A + C = 3 \) (4) \( A + 3 + C = 6 \implies A + C = 3 \) (5) Substitute \( B = 3 \) into (3): \( 9A - 3(3) + C = -2 \implies 9A + C = 7 \) (6) Subtract (4) from (6): \( (9A + C) - (A + C) = 7 - 3 \implies 8A = 4 \implies A = \frac{1}{2} \) Substitute \( A = \frac{1}{2} \) into (4): \( \frac{1}{2} + C = 3 \implies C = 3 - \frac{1}{2} = \frac{5}{2} \) The equation of the parabola is \( x = \frac{1}{2}y^2 + 3y + \frac{5}{2} \) The parabola cuts the X-axis when \( y = 0 \).
\( x = \frac{1}{2}(0)^2 + 3(0) + \frac{5}{2} = \frac{5}{2} \) The point at which the parabola cuts the X-axis is \( \left( \frac{5}{2}, 0 \right) \).