$${e^2} = 1 - {{{b^2}} \over {{a^2}}} = 1 - {5 \over 9} = {4 \over 9}$$
The foci are ($$\pm$$ ae, 0) i.e. (2, 0) and ($$-$$2, 0).
The parabola P1 is $${y^2} = 8x$$ and P2 is $${y^2} = - 16x$$
As tangent with slope m1 to P1 passes through ($$-$$4, 0), we have
$$y = {m_1}x + {2 \over {{m_1}}}$$ giving $$0 = - 4{m_1} + {2 \over {{m_1}}}$$
i.e. $$4m_1^2 = 2 \Rightarrow m_1^2 = {1 \over 2}$$
Again for tangent with slope m2 to P2 passing through (2, 0), we have
$$y = {m_2}x - {4 \over {{m_2}}} \Rightarrow 0 = 2{m_2} - {4 \over {{m_2}}}$$
$$ \Rightarrow 2m_2^2 = 4$$ $$\therefore$$ $$m_2^2 = 2$$
Thus, $${1 \over {m_1^2}} + m_2^2 = 2 + 2 = 4$$
Let, P(t2, 2t) be any point on the parabola y2 = 4x. C be the mirror image of the parabola y2 = 4x with respect to the line UV : x + y + 4 = 0.
The curve C cuts the line KL : y = $$-$$5 at A and B.
Let, B($$\alpha$$, $$\beta$$) be the image of the point P(t2, 2t).
Clearly, PB $$\bot$$ UV and PQ = QB.
$$\therefore$$ $${{\alpha - {t^2}} \over {\beta - 2t}} \times ( - 1) = - 1$$
or, $$\alpha - {t^2} = \beta - 2t$$ ...... (1)
The point of intersection of the lines UV and KL is R.
Let us join P and R.
From $$\Delta$$PQR and $$\Delta$$BQR,
(i) BQ = PQ [$$\because$$ B is the image of P]
(ii) $$\angle$$PQR = $$\angle$$RQB = 90$$^\circ$$ [$$\because$$ PB $$\bot$$ UV]
(iii) QR common
$$\therefore$$ $$\Delta$$PQR $$ \cong $$ $$\Delta$$BQR [by SAS congruence criterion]
$$\therefore$$ $$\angle$$QRP = $$\angle$$BRQ [CPCT]
$$\because$$ slope of x + y + 4 = 0 is $$-$$1,
$$\therefore$$ $$\angle$$UTO = 135$$^\circ$$
$$\therefore$$ $$\angle$$OTR = 45$$^\circ$$
Again, X'X || KL and UV transversal.
$$\therefore$$ $$\angle$$OTR = $$\angle$$TRB = 45$$^\circ$$ $$\therefore$$ $$\angle$$BRQ = $$\angle$$QRP = 45$$^\circ$$
$$\therefore$$ $$\angle$$PRB = 90$$^\circ$$ $$\therefore$$ PR $$\bot$$ KL
$$\therefore$$ coordinates of R are (t2, $$\beta$$).
$$\because$$ the point R lies on KL,
$$\therefore$$ $$\beta$$ = $$-$$5
Again, the point R lies on the straight line x + y + 4 = 0.
$$\therefore$$ t2 + $$\beta$$ + 4 = 0
or, t2 $$-$$ 5 + 4 = 0 [$$\because$$ $$\beta$$ = $$-$$5]
or, t2 = 1 or, t = $$\pm$$ 1
when t = 1, $$\beta$$ = $$-$$5, then (1) $$\Rightarrow$$ $$\alpha$$ $$-$$ 1 = $$-$$ 5 $$-$$ 2 or $$\alpha$$ = $$-$$6
when, t = $$-$$1, $$\beta$$ = $$-$$5, then (1) $$\Rightarrow$$ $$\alpha$$ $$-$$ 1 = $$-$$ 5 + 2 or, $$\alpha$$ = $$-$$2
So, the coordinates of A and B are ($$-$$6, $$-$$5) and ($$-$$2, $$-$$5) respectively.
$$\therefore$$ AB = 4 units
So, the distance between A and B is 4 units.
