Question

locus of mid point of chord of parabola which are at a constant lenght 2l

Answer 1

$(4ax - y^2)(y^2 + 4a^2) = 4a^2l^2$

Answer 2

Let the equation of the parabola be $y^2 = 4ax$. Let the two endpoints of a chord be $A(at_1^2, 2at_1)$ and $B(at_2^2, 2at_2)$. The length of the chord $AB$ is given as $2l$.

Using the distance formula, the square of the length of the chord $AB$ is: $(2l)^2 = (at_1^2 - at_2^2)^2 + (2at_1 - 2at_2)^2$ $4l^2 = a^2(t_1^2 - t_2^2)^2 + 4a^2(t_1 - t_2)^2$ $4l^2 = a^2[(t_1 - t_2)(t_1 + t_2)]^2 + 4a^2(t_1 - t_2)^2$ $4l^2 = a^2(t_1 - t_2)^2(t_1 + t_2)^2 + 4a^2(t_1 - t_2)^2$ $4l^2 = a^2(t_1 - t_2)^2[(t_1 + t_2)^2 + 4]$ --- (1)

Let $(h, k)$ be the midpoint of the chord $AB$. Using the midpoint formula: $h = \frac{at_1^2 + at_2^2}{2} = \frac{a}{2}(t_1^2 + t_2^2)$ --- (2) $k = \frac{2at_1 + 2at_2}{2} = a(t_1 + t_2)$ --- (3)

From (3), we get $t_1 + t_2 = \frac{k}{a}$ --- (4) From (2), we get $t_1^2 + t_2^2 = \frac{2h}{a}$ --- (5)

We know the identity $(t_1 - t_2)^2 = (t_1 + t_2)^2 - 4t_1t_2$. Also, $2t_1t_2 = (t_1+t_2)^2 - (t_1^2+t_2^2)$. Substituting (4) and (5) into the expression for $2t_1t_2$: $2t_1t_2 = \left(\frac{k}{a}\right)^2 - \frac{2h}{a} = \frac{k^2}{a^2} - \frac{2h}{a}$

Now substitute $t_1+t_2$ and $2t_1t_2$ into the identity for $(t_1 - t_2)^2$: $(t_1 - t_2)^2 = \left(\frac{k}{a}\right)^2 - 2\left(\frac{k^2}{a^2} - \frac{2h}{a}\right)$ $(t_1 - t_2)^2 = \frac{k^2}{a^2} - \frac{2k^2}{a^2} + \frac{4h}{a}$ $(t_1 - t_2)^2 = \frac{4h}{a} - \frac{k^2}{a^2}$ --- (6)

Now, substitute (4) and (6) into equation (1): $4l^2 = a^2 \left( \frac{4h}{a} - \frac{k^2}{a^2} \right) \left[ \left( \frac{k}{a} \right)^2 + 4 \right]$ $4l^2 = a^2 \left( \frac{4ah - k^2}{a^2} \right) \left( \frac{k^2 + 4a^2}{a^2} \right)$ $4l^2 = \frac{(4ah - k^2)(k^2 + 4a^2)}{a^2}$

Multiply both sides by $a^2$: $4a^2l^2 = (4ah - k^2)(k^2 + 4a^2)$

To find the locus, replace $(h, k)$ with $(x, y)$: $4a^2l^2 = (4ax - y^2)(y^2 + 4a^2)$

This is the required locus. It can also be expanded as: $4a^2l^2 = 4axy^2 + 16a^3x - y^4 - 4a^2y^2$ Rearranging the terms: $y^4 - 4axy^2 + 4a^2y^2 - 16a^3x + 4a^2l^2 = 0$

The final answer is $(4ax - y^2)(y^2 + 4a^2) = 4a^2l^2$.