A tangent and a normal are drawn at the point P ( 2, - 4 ) on the...

Equation of tangent at $\left(2, - 4\right)\left(T = 0\right)$

$- 4y = 4\left(x + 2\right)$

$x + y + 2 = 0 . . . ( 1 )$

Equation of normal

$x - y + \lambda =0$

\left(2, - 4\right)$</p><p>$\lambda = - 6$</p><p>Thus,</p><p>$x - y = 6 . . . ( 2 )$</p><p>Equation of normal</p><p>$POI$of$( 1 ) \text{ and } x = - 2$is$A\left(- 2, 0\right)$</p><p>$POI$of \left(2\right) \text{ and } x = - 2$ is $B\left(- 2, - 8\right)$

Given

$AQBP$ is a square

${m}_{AQ} \cdot {m}_{AP} = - 1$

$\dfrac{b}{a + 2} \times \dfrac{4}{- 4} = - 1$

$a + 2 = b$

Also $PQ$ must be parallel to $x -$axis

thus,

$b = - 4$

$a = - 6$

Thus, $2a + b = - 16$