Question

Find the locus of the mid-points of the chords of the parabola y²=4ax which subtend a right angle at vertex of the parabola.

Answer 1

The locus of midpoints of chords subtending a right angle at the vertex of the parabola
`\(y^2 = 4ax\)` is
`\(y = \pm √(a)`, forming two lines parallel to the x-axis.

Let's consider the parabola
`\(y^2 = 4ax\)`. The equation of a chord joining two points
`\((x_1, y_1)\)` and
`\((x_2, y_2)\)` on the parabola is given by:

`\[ y - (y_1 + y_2)/(2) = m(x - (x_1 + x_2)/(2)) \]`

where m is the slope of the chord.

Given that the chord subtends a right angle at the vertex of the parabola, the product of the slopes of the two chords is -1. Therefore:

`\[ m_1 \cdot m_2 = -1 \]`

Now, let's find the slopes of the chords on the parabola
`\(y^2 = 4ax\)`.

The equation
`\(y^2 = 4ax\)` implies
`\(y = \pm √(4ax)\)`.

1. For the first chord, let
`\(y_1 = √(4ax_1)\)` and
`\(y_2 = -√(4ax_2)\)`, where
`\(x_1\)` and
`\(x_2\)` are the x-coordinates of the two points on the parabola.

Slope
`\(m_1\)` for the first chord:

`\[ m_1 = (-√(4ax_2) - √(4ax_1))/(x_2 - x_1) \]`

2. For the second chord, let
`\(y_1 = -√(4ax_1)\)` and
`\(y_2 = √(4ax_2)\)`.

Slope
`\(m_2\)` for the second chord:

`\[ m_2 = (√(4ax_2) + √(4ax_1))/(x_2 - x_1) \]`

Now, the product of the slopes:

`\[ m_1 \cdot m_2 = (-√(4ax_2) - √(4ax_1))/(x_2 - x_1) \cdot (√(4ax_2) + √(4ax_1))/(x_2 - x_1) \]\[ m_1 \cdot m_2 = (-4ax_2 + 4ax_1)/((x_2 - x_1)^2) \]`

To have
`\(m_1 \cdot m_2 = -1\)`, we need:

`\[ x_2 - x_1 = \pm √(a) \]`

Now, the locus of the midpoints
`\((h, k)\)` of the chords is given by:

`\[ (h, k) = \left((x_1 + x_2)/(2), (y_1 + y_2)/(2)\right) \]`

Substituting the values of
`\(x_1\)` and
`\(x_2\)` we found above, we get the locus of midpoints as:

`\[ (h, k) = (h, \pm (1)/(2) √(4ah)) \]`

Thus, the locus of the midpoints of the chords subtending a right angle at the vertex is
`\(y = \pm √(a) \)`.