QJEEAdv15Mad9a5: JEE Advanced 2015 Mathematics - Class 11 JEEAdv15Mad9a5 Solution: Let the curve be the mirror image of the parabola with respect…

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.