Meany straight lines ||

A=Arithmetic mean of a & b, G=geometric mean of a & b, H=harmonic mean of a & b.

We know that , $G=\sqrt{AH}$

Given equation,

$\implies Ax^2+Hy^2+2Gxy=32$

$\implies Ax^2+Hy^2+2\sqrt{AH}xy=32$ . …(because $G=\sqrt{AH}$ )

$\implies (\sqrt{A}x)^2+(\sqrt{H}y)^2+2\sqrt{A}\sqrt{H}xy=32$ ……… (this is in form of $x^2+y^2+2xy=(x+y)^2$ )

$\implies (\sqrt{A}x+\sqrt{H}y)^2=32$

$\implies \sqrt{A}x+\sqrt{H}y=\pm\sqrt{32}$

$\implies y=-\sqrt{\frac{A}{H}}x\pm\sqrt{\frac{32}{H}}$

This is in form of $y=mx+c$

So, the intercepts are $+\sqrt{\frac{32}{H}}$ and $-\sqrt{\frac{32}{H}}$

Given, H=2

Y- intercept points are (0,4) and (0,-4)

Distance between the intercepts is $\boxed{8}$

Let $a=b$ (Considering the equality case)

$A=\dfrac{2a}{2}=a$

$G=\sqrt{a^2}=a$

$H=\dfrac{2a^2}{2a}=a$

But given that $H=2$

$\implies A=G=H=2$

Hence, the combined equation becomes $2x^2+2y^2+4xy=32$ .

Which is the same as writing $x^2+y^2+2xy=16$

$\implies (x+y)^2=16$

$\implies (x+y)=\pm4$

Therefore, the two separated equations are $x+y=4$ and $x+y=-4$ .

Clearly intercepts are $(0,4)$ and $(0,-4)$ .

Distance between them, $d$ = $\boxed{8}$