Question

consider the feet of normals from (3,5) to y^2=8x. arithmetic mean of ordinates of these points are?

Answer 1

0

Answer 2

The equation of the parabola is $y^2 = 8x$. This is in the form $y^2 = 4ax$, where $4a = 8$, so $a = 2$. Let the point from which the normals are drawn be $(x_1, y_1) = (3, 5)$.

The equation of the normal to the parabola $y^2 = 4ax$ at the point $(at^2, 2at)$ is given by $y = -tx + 2at + at^3$. If this normal passes through the point $(x_1, y_1)$, then substituting $(x_1, y_1)$ into the equation gives: $y_1 = -tx_1 + 2at + at^3$ Rearranging this equation into a cubic in $t$: $at^3 + (2a - x_1)t - y_1 = 0$

Substitute the values $a = 2$, $x_1 = 3$, and $y_1 = 5$: $2t^3 + (2(2) - 3)t - 5 = 0$ $2t^3 + (4 - 3)t - 5 = 0$ $2t^3 + t - 5 = 0$

This cubic equation gives the values of the parameter $t$ for the feet of the normals drawn from $(3, 5)$ to the parabola. Let the roots of this equation be $t_1, t_2, t_3$. The feet of the normals are the points $(at_i^2, 2at_i)$ for $i=1, 2, 3$. The ordinates of these points are $y'_i = 2at_i$. We are asked to find the arithmetic mean of these ordinates, which is $(y'_1 + y'_2 + y'_3)/3$. Arithmetic mean = $(2at_1 + 2at_2 + 2at_3)/3 = 2a(t_1 + t_2 + t_3)/3$.

For a cubic equation $At^3 + Bt^2 + Ct + D = 0$, the sum of the roots is $t_1 + t_2 + t_3 = -B/A$. In the equation $2t^3 + 0t^2 + 1t - 5 = 0$, we have $A = 2$, $B = 0$, $C = 1$, and $D = -5$. The sum of the roots is $t_1 + t_2 + t_3 = -0/2 = 0$.

Now, substitute the values of $a$ and $(t_1 + t_2 + t_3)$ into the expression for the arithmetic mean of the ordinates: Arithmetic mean = $2a(t_1 + t_2 + t_3)/3 = 2(2)(0)/3 = 0$.

Alternatively, we can find the cubic equation directly in terms of the ordinate $y'$. Let the ordinate of a point on the parabola be $y'$. Since $y^2 = 8x$, we have $x = y'^2/8$. The parameter $t$ is related to the ordinate by $y' = 2at = 2(2)t = 4t$, so $t = y'/4$. Substitute $x = y'^2/8$, $y = y'$, $a=2$, $x_1=3$, $y_1=5$ into the normal equation $y - y_1 = -t(x - x_1)$: $y' - 5 = -(y'/4)(y'^2/8 - 3)$ $y' - 5 = -(y'/4)((y'^2 - 24)/8)$ $y' - 5 = -y'(y'^2 - 24)/32$ $32(y' - 5) = -y'(y'^2 - 24)$ $32y' - 160 = -y'^3 + 24y'$ $y'^3 + 32y' - 24y' - 160 = 0$ $y'^3 + 8y' - 160 = 0$

This is a cubic equation in $y'$. Let the roots (the ordinates of the feet of the normals) be $y'_1, y'_2, y'_3$. The equation is $1 \cdot y'^3 + 0 \cdot y'^2 + 8 \cdot y' - 160 = 0$. The sum of the roots is $y'_1 + y'_2 + y'_3 = -(\text{coefficient of } y'^2) / (\text{coefficient of } y'^3) = -0/1 = 0$. The arithmetic mean of the ordinates is $(y'_1 + y'_2 + y'_3)/3 = 0/3 = 0$.

Note: While the point (3, 5) is such that only one real normal can be drawn to the parabola $y^2 = 8x$, the question asks for the arithmetic mean of the ordinates of "the feet of normals". In such problems, we consider the roots of the cubic equation, which correspond to the feet of the normals (including complex ones if the point is not in the correct region). The sum of the roots property holds for all roots, real or complex.