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) and B(2t, 0)

1. Midpoint of AB: 
Using the midpoint formula:
Midpoint M = $\left( \frac{0 + 2t}{2}, \frac{4 + 0}{2} \right) = (t, 2)$

2. Slope of AB:
$m_1 = \frac{0 - 4}{2t - 0} = \frac{-4}{2t} = \frac{-2}{t}$
Let $m_2$ be the slope of the perpendicular bisector. Then:
$m_1 \cdot m_2 = -1 \Rightarrow \left( \frac{-2}{t} \right) \cdot m_2 = -1$
$\Rightarrow m_2 = \frac{t}{2}$

3. Equation of the perpendicular bisector:
Using point-slope form with point M(t, 2) and slope $m_2 = \frac{t}{2}$:
$y - 2 = \frac{t}{2}(x - t)$

4. Intersection with y-axis (put $x = 0$):
$y - 2 = \frac{t}{2}(0 - t) = \frac{-t^2}{2}$
$y = 2 - \frac{t^2}{2} = \frac{4 - t^2}{2}$
So, point R = $\left( 0, \frac{4 - t^2}{2} \right)$

5. Midpoint of MR (call it point P):
M = $(t, 2)$, R = $\left( 0, \frac{4 - t^2}{2} \right)$
Midpoint P = $\left( \frac{t + 0}{2}, \frac{2 + \frac{4 - t^2}{2}}{2} \right)$
$\Rightarrow P = \left( \frac{t}{2}, \frac{8 - t^2}{4} \right)$

6. Eliminate parameter t:
Let $P = (h, k)$, so $h = \frac{t}{2} \Rightarrow t = 2h$
Substitute in $k = \frac{8 - t^2}{4}$:
$k = \frac{8 - (2h)^2}{4} = \frac{8 - 4h^2}{4} = 2 - h^2$

Final locus equation:
$k = 2 - h^2$
This is the equation of a parabola.

Answer: Option (B) : a parabola

Answer 2

Given:

A = (0, 4) 

B = (2t, 0)

L is the midpoint of AB.

The perpendicular bisector of AB meets the y-axis at M.

N is the midpoint of LM.

We want to find the locus of N.

Step 1: Find the coordinates of L.

L is the midpoint of AB, so its coordinates are:

\(L = \left( \frac{0 + 2t}{2}, \frac{4 + 0}{2} \right) = (t, 2)\)

Step 2: Find the equation of the perpendicular bisector of AB.

The slope of AB is:

\(m_{AB} = \frac{0 - 4}{2t - 0} = \frac{-4}{2t} = -\frac{2}{t}\)

The slope of the perpendicular bisector is the negative reciprocal of \(m_{AB}\):

\(m_{\perp} = \frac{t}{2}\)

The perpendicular bisector passes through the midpoint L (t, 2). Using the point-slope form of a line:

\(y - 2 = \frac{t}{2}(x - t)\)

Step 3: Find the coordinates of M.

M is the point where the perpendicular bisector intersects the y-axis, so x = 0. Substitute x = 0 into the equation of the perpendicular bisector:

\(y - 2 = \frac{t}{2}(0 - t)\)

\(y = 2 - \frac{t^2}{2}\)

So, M = \((0, 2 - \frac{t^2}{2})\)

Step 4: Find the coordinates of N.

N is the midpoint of LM. L = (t, 2) and M = \((0, 2 - \frac{t^2}{2})\). Therefore, the coordinates of N are:

\(N = \left( \frac{t + 0}{2}, \frac{2 + 2 - \frac{t^2}{2}}{2} \right) = \left( \frac{t}{2}, \frac{4 - \frac{t^2}{2}}{2} \right) = \left( \frac{t}{2}, 2 - \frac{t^2}{4} \right)\)

Step 5: Find the locus of N.

Let the coordinates of N be (x, y). Then:

\(x = \frac{t}{2} \implies t = 2x\)

\(y = 2 - \frac{t^2}{4}\)

Substitute \(t = 2x\) into the equation for y:

\(y = 2 - \frac{(2x)^2}{4} = 2 - \frac{4x^2}{4} = 2 - x^2\)

\(y = 2 - x^2\)

\(x^2 = 2 - y\)

This is the equation of a parabola.