TMUA practice: Straight Lines, problem 4

Write $M = (0, m)$ and $N = (0, n)$, with $m < n$ (as $M$ is below $N$) and $n - m = 4$.

Line $AM$ has $x$-intercept $-4$ and $y$-intercept $m$, so in intercept form $\dfrac{x}{-4} + \dfrac{y}{m} = 1$. For the point $P = (x, y)$ on it, $\dfrac{y}{m} = 1 + \dfrac{x}{4}$, hence $m = \dfrac{4y}{4 + x}$.

Line $BN$ has $x$-intercept $4$ and $y$-intercept $n$, so $\dfrac{x}{4} + \dfrac{y}{n} = 1$, giving $n = \dfrac{4y}{4 - x}$.

Applying $n - m = 4$:

$\frac{4y}{4 - x} - \frac{4y}{4 + x} = 4.$

Combine the fractions:

So $8xy = 4(16 - x^2)$, that is $2xy = 16 - x^2$, which rearranges to

$x^2 + 2xy = 16.$

Answer: B.