Question

Show that the normal to the rectangular hyperbola xy = 52 at the point t meets the curve again at a point t' such that t²t' = -1.

Answer 1

The question as stated is likely incorrect. The correct relation is t³t' = -1.

Answer 2

The question asks to show that the normal to the rectangular hyperbola $xy = 52$ at the point $t$ meets the curve again at a point $t'$ such that $t^2t' = -1$. However, the standard derivation using the parametric representation $x = \sqrt{52}t$, $y = \sqrt{52}/t$ leads to the result $t^3t' = -1$.

Derivation:

1. The slope of the tangent at $P(\sqrt{52}t, \sqrt{52}/t)$ is $m_t = -1/t^2$.
2. The slope of the normal at $P$ is $m_n = t^2$.
3. The equation of the normal at $P$ is $y - \sqrt{52}/t = t^2(x - \sqrt{52}t)$, which simplifies to $t^3x - ty + \sqrt{52}(1 - t^4) = 0$.
4. Let the normal intersect the hyperbola again at $Q(\sqrt{52}t', \sqrt{52}/t')$. Substitute the coordinates of $Q$ into the normal equation: $t^3(\sqrt{52}t') - t(\sqrt{52}/t') + \sqrt{52}(1 - t^4) = 0$.
5. Divide by $\sqrt{52}$: $t^3t' - t/t' + (1 - t^4) = 0$.
6. Multiply by $t'$: $t^3(t')^2 + (1 - t^4)t' - t = 0$.
7. This is a quadratic equation in $t'$. The product of the roots is $t \cdot t' = -t/t^3 = -1/t^2$.
8. Therefore, $t t' = -1/t^2$, which implies $t^3 t' = -1$.

This standard derivation shows that the normal to the rectangular hyperbola $xy = 52$ at the point $t$ meets the curve again at a point $t'$ such that $t^3t' = -1$. The question asks to show $t^2t' = -1$. This is inconsistent with the derived result. It is highly probable that the question contains a typo.

If there is a specific, non-standard definition of 'point t' or 'point t'' that leads to the relation $t^2t' = -1$, it is not provided in the question. Without such a definition, it is not possible to rigorously show that $t^2t' = -1$ holds for any point $t$ on the hyperbola.