Find the range of values of $k$ for which the line $y = kx + 3$ lies entirely below the curve $xy + 20 = 5y$ for all real values of $x$. The quadratic equation $mx^2 + 5nx + 9m = 0$ has equal roots. Given that $m$ and $n$ are positive, find $\frac{m}{n}$. Determine also the range of values of $\frac{m}{n}$ for which the quadratic equation has real roots.

1. Substitute the line into the curve equation

Substitute $y = kx + 3$ into $xy + 20 = 5y$:
$x(kx+3) + 20 = 5(kx+3)$
$kx^2 + 3x + 20 = 5kx + 15$

2. Form the quadratic equation in standard form

$kx^2 + (3 - 5k)x + 5 = 0$
This is the equation whose roots are the $x$-coordinates of intersection of the line and curve.

3. Apply the no-intersection (discriminant) condition

For the line to lie entirely below the curve (never meeting it) for all real $x$, the quadratic must have no real roots:
$b^2 - 4ac < 0$
$(3 - 5k)^2 - 4(k)(5) < 0$

4. Expand and simplify the inequality

$9 - 30k + 25k^2 - 20k < 0$
$25k^2 - 50k + 9 < 0$

5. Solve the quadratic inequality in k

Factorise: $(5k - 1)(5k - 9) < 0$
$\frac{1}{5} < k < \frac{9}{5}$

1. Set up the discriminant equation for equal roots

For equal roots, the discriminant $b^2 - 4ac = 0$.
$(5n)^2 - 4(m)(9m) = 0$
$25n^2 - 36m^2 = 0$

2. Solve for the ratio $\frac{m}{n}$

$36m^2 = 25n^2$
$\frac{m^2}{n^2} = \frac{25}{36}$
Since $m$ and $n$ are positive, $\frac{m}{n} > 0$.
$\frac{m}{n} = \sqrt{\frac{25}{36}} = \frac{5}{6}$

3. Set up the discriminant inequality for real roots

For real roots, the discriminant $b^2 - 4ac \geqslant 0$.
$25n^2 - 36m^2 \geqslant 0$
$36m^2 \leqslant 25n^2$
$\frac{m^2}{n^2} \leqslant \frac{25}{36}$

4. Solve for the range of $\frac{m}{n}$

$-\frac{5}{6} \leqslant \frac{m}{n} \leqslant \frac{5}{6}$
Since $m$ and $n$ are given to be positive, $\frac{m}{n} > 0$.
Therefore, $0 < \frac{m}{n} \leqslant \frac{5}{6}$.