Find the angle at which the curves x=sqrt5 cosh t

Find the angle at which the curves $x=\sqrt5\cosh t,\ y=\sqrt5\sinh t$ and $x=3\sqrt2\cos t,\ y=2\sqrt2\sin t$ intersect.

Step 1: Identify the two curves

First: using $\cosh^{2}t-\sinh^{2}t = 1$,

\[ x^{2}-y^{2} = 5\bigl(\cosh^{2}t-\sinh^{2}t\bigr) = 5, \]

the right branch of the hyperbola $x^{2}-y^{2} = 5$ (right, because $\cosh t \gt 0$).

Second: using $\cos^{2}t+\sin^{2}t = 1$,

\[ \frac{x^{2}}{18}+\frac{y^{2}}{8} = 1. \]

Step 2: Find the intersection points

From the hyperbola $y^{2} = x^{2}-5$; substituting into the ellipse and multiplying by $72$: \[\frac{x^{2}}{18}+\frac{x^{2}-5}{8} = 1 \;\Longrightarrow\; 4x^{2}+9\bigl(x^{2}-5\bigr) = 72 \;\Longrightarrow\; 13x^{2} = 117 \;\Longrightarrow\; x^{2} = 9.\]

Only $x = 3$ lies on the right branch, and then $y^{2} = 4$: $P_{1}(3,\,2), \qquad P_{2}(3,\,-2)$.

Step 3: Slopes at $P_{1}(3,2)$

Hyperbola: $2x-2y\,y' = 0$, so

\[ k_{1} = \frac{x}{y} = \frac{3}{2}. \]

Ellipse: $\dfrac{2x}{18}+\dfrac{2y}{8}y' = 0$, so

\[ k_{2} = -\frac{4x}{9y} = -\frac{12}{18} = -\frac{2}{3}. \]

Step 4: Test for orthogonality

\[ k_{1}k_{2} = \frac{3}{2}\cdot\left(-\frac{2}{3}\right) = -1. \]

The product is exactly $-1$, so the tangents are perpendicular. (The angle formula would give $\tan\alpha = \left|\dfrac{k_{1}-k_{2}}{1+k_{1}k_{2}}\right|$ with a vanishing denominator — the analytic signal of a right angle.)

Step 5: The second point

At $P_{2}(3,-2)$ both slopes change sign, $k_{1} = -\tfrac32$ and $k_{2} = \tfrac23$, so again $k_{1}k_{2} = -1$.

Step 6: Why it had to happen

In general the two slopes are $k_{1} = \dfrac{x}{y}$ and $k_{2} = -\dfrac{4x}{9y}$, whose product is $-\dfrac{4x^{2}}{9y^{2}}$. At the intersection $x^{2} = 9$, $y^{2} = 4$, so

\[k_{1}k_{2} = -\frac{4\cdot 9}{9\cdot 4} = -1 \;\;\checkmark\]

\[\text{Both conics share the foci } (\pm\sqrt{10},0): \quad 18-8 = 10 \quad\text{and}\quad 5+5 = 10 \;\;\checkmark\]