Tangents to Parabola and Properties of Focus Triangle

The given equation of the parabola $E$ is $y^2 = 8x$ Comparing this with the standard equation $y^2 = 4ax$, we get $a = 2$.

From this, we find:

• Focus ($F$): $(a, 0) = (2, 0)$
• Directrix: $x = -a \implies x = -2$

The given point is $P = (-2, 4)$. Notice that $P$ lies on the directrix $x = -2$.

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Analysis of Statements:

Statement A: The triangle $PFQ$ forms a right-angled triangle

The equation of a tangent to the parabola $y^2 = 8x$ at any point $Q(2t^2, 4t)$ is given by: $ty = x + 2t^2$

Since this tangent passes through $P(-2, 4)$, we substitute $x = -2$ and $y = 4$: $4t = -2 + 2t^2 \implies t^2 - 2t - 1 = 0 \implies t^2 - 1 = 2t$

Now, let us calculate the slope of the line joining focus $F(2, 0)$ and point $Q(2t^2, 4t)$: $m_{FQ} = \frac{4t - 0}{2t^2 - 2} = \frac{4t}{2(t^2 - 1)} = \frac{2t}{t^2 - 1}$

Using $t^2 - 1 = 2t$, we get: $m_{FQ} = \frac{2t}{2t} = 1$

Next, we calculate the slope of the line joining $P(-2, 4)$ and $F(2, 0)$: $m_{PF} = \frac{4 - 0}{-2 - 2} = \frac{4}{-4} = -1$

Since $m_{FQ} \cdot m_{PF} = (1)(-1) = -1$, the line segment $PF$ is perpendicular to $FQ$ (i.e., $\angle PFQ = 90^\circ$).

Thus, $\triangle PFQ$ is a right-angled triangle at $F$.
Statement A is TRUE.

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Statement B: The triangle $QPQ'$ forms a right-angled triangle

Since point $P(-2, 4)$ lies on the directrix ($x = -2$), the pair of tangents $PQ$ and $PQ'$ drawn from $P$ to the parabola are perpendicular to each other.

Therefore, $\angle QPQ' = 90^\circ$, making $\triangle QPQ'$ a right-angled triangle.
Statement B is TRUE.

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Statement C: The distance between points $P$ and $F$ equals $5\sqrt{2}$

Using the distance formula between $P(-2, 4)$ and $F(2, 0)$: $PF = \sqrt{(2 - (-2))^2 + (0 - 4)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$

Since $PF = 4\sqrt{2} \neq 5\sqrt{2}$,
Statement C is FALSE.

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Statement D: The focus $F$ lies on the line segment joining $Q$ and $Q'$

The line segment $QQ'$ is the chord of contact of tangents drawn from $P(-2, 4)$ to the parabola $y^2 = 8x$.
The equation of the chord of contact is $T = 0$: $y \cdot 4 = 4(x - 2) \implies 4y = 4x - 8 \implies x - y - 2 = 0$

Substituting the coordinates of focus $F(2, 0)$ into the line equation: $2 - 0 - 2 = 0$

Since the coordinates of $F$ satisfy the equation of the chord of contact $QQ'$, the focus $F$ lies on the chord $QQ'$.
Statement D is TRUE.

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Conclusion:

The correct options are A, B, and D.