Question

If the normals at $(x_i,y_i):i=1,2,3,4$ on the rectangular hyperbola $xy=c^2$ meet at the point $(\alpha,\beta)$

• A. $c\beta$
• B. $c\alpha$
• C. $\alpha$
• D. $\beta$

Answer 1

Answer: (C)

$\alpha$

Answer 2

The equation of the rectangular hyperbola is $xy=c^2$. A point on the hyperbola can be parameterized as $(ct, c/t)$. The slope of the tangent at $(ct, c/t)$ is $dy/dx = -c^2/x^2 = -c^2/(ct)^2 = -1/t^2$. The slope of the normal at $(ct, c/t)$ is $m_n = -1/(-1/t^2) = t^2$. The equation of the normal at $(ct, c/t)$ is $y - c/t = t^2(x - ct)$. $y - c/t = t^2x - ct^3$ $t^2x - y + c/t - ct^3 = 0$ Multiplying by $t$, we get $t^3x - ty + c - ct^4 = 0$. Rearranging, we get $ct^4 - t^3x + ty - c = 0$.

Let the four points where the normals meet at $(\alpha, \beta)$ be $(x_i, y_i)$ for $i=1,2,3,4$. These points can be written as $(ct_i, c/t_i)$, where $t_i$ are the roots of the equation obtained by substituting $(\alpha, \beta)$ into the normal equation: $ct^4 - \alpha t^3 + \beta t - c = 0$. This is a quartic equation in $t$. Let the roots be $t_1, t_2, t_3, t_4$. According to Vieta's formulas: Sum of roots: $\sum t_i = t_1+t_2+t_3+t_4 = -(-\alpha)/c = \alpha/c$.

We have $x_i = ct_i$ and $y_i = c/t_i$ for $i=1,2,3,4$.

The value of $\sum x_i$: $\sum x_i = x_1+x_2+x_3+x_4 = ct_1+ct_2+ct_3+ct_4 = c(t_1+t_2+t_3+t_4) = c(\alpha/c) = \alpha$.