Question

Let A be the point (0,4) in the xy-plane and B be the point (2t,0). Let L be AB's midpoint and let AB's perpendicular bisector meet the y-axis M. Let N be the midpoint of LM. The locus of N is

• A. a circle
• B. a parabola
• C. a straight line
• D. a hyperbola

Answer 1

The Correct Option is B

Given points:
A(0, 4)
B(2t, 0)

1. Find the coordinates of the midpoint M of AB:
Using the midpoint formula, the coordinates of M are (t, 2).

2. Determine the slope of AB (m1) and the slope of the perpendicular bisector (m2):
The slope of AB (m1) = (0 - 4) / (2t - 0) = -2/t.

Since the product of the slopes of perpendicular lines is -1, we have:
m1 * m2 = -1
(-2/t) * m2 = -1
m2 = t/2.

3. Write the equation of the perpendicular bisector using point-slope form:
Using the point-slope form (y - y1) = m(x - x1) and the point M(t, 2):
y - 2 = (t/2)(x - t)

4. Find the coordinates of the point where the perpendicular bisector intersects the y-axis (let's call this point R):
Substitute x = 0 into the equation:
y - 2 = (t/2)(0 - t)
y = 2 - t^2 / 2
y = (4 - t^2) / 2

So, the coordinates of point R are (0, (4 - t^2) / 2).

5. Find the midpoint coordinates of MR (let's call this point P):
Using the midpoint formula:
Coordinates of P: (t/2, (2 + (4 - t^2) / 2) / 2)
Simplifying, we get P: (t/2, (8 - t^2) / 4).

6. Eliminate the parameter t to find the locus of point P:
Let the coordinates of P be (h, k).
From the earlier derived equations, we have:
h = t/2, k = (8 - t^2) / 4
Squaring both sides of h = t/2: h^2 = t^2 / 4
Replacing t^2 with 4h^2 in k: k = 2 - h^2

The correct answer is option (B) : a parabola