Answer: \(P=(\sqrt6,\frac{\sqrt6}{3})\), \(Q=(-\sqrt6,-\frac{\sqrt6}{3})\); \(PQ=\frac{4\sqrt{15}}{3}\).

We start with the main method. Substitute the line into the curve; a tangent corresponds to a repeated root, so the discriminant is zero.

(a) Substitute \(y=\frac2x\) into

\(4x^2-3y^2+xy=24.\)

This gives

\(4x^2-3\left(\frac2x\right)^2+x\left(\frac2x\right)=24.\)

So

\(4x^2-\frac{12}{x^2}+2=24.\)

Multiply by \(x^2\):

\(4x^4-12+2x^2=24x^2.\)

Hence

\(4x^4-22x^2-12=0,\)

or

\(2x^4-11x^2-6=0.\)

Let \(u=x^2\). Then

\(2u^2-11u-6=0.\)

Factorise:

\((2u+1)(u-6)=0.\)

Since \(u=x^2\geq0\), we take \(u=6\). Thus

\(x=\pm\sqrt6.\)

Using \(y=\frac2x\), the points are

\(\left(\sqrt6,\frac{\sqrt6}{3}\right)\quad\text{and}\quad \left(-\sqrt6,-\frac{\sqrt6}{3}\right).\)

(b) The distance between these two points is

\(PQ=\sqrt{(2\sqrt6)^2+\left(\frac{2\sqrt6}{3}\right)^2}.\)

So

\(PQ=\sqrt{24+\frac83}=\sqrt{\frac{80}{3}} =\frac{4\sqrt{15}}{3}.\)

This completes the solution and gives the required result.