Let the tangent to the curve x ^ 2 + 2x - 4y + 9 = 0 at the point...

Given,

The tangent to the curve ${x}^{2} + 2x - 4y + 9 = 0$ at the point $P\left(1, 3\right)$ on it meet the $y -$axis at $A$,

So, tangent to circle at point $P\left(1, 3\right)$ is given by,

$x + \left(x + 1\right) - 2\left(y + 3\right) + 9 = 0$

$⇒ 2x - 2y = - 4$

So, on $y -$axis, $A\left(0, 2\right)$

And the line passing through $P$ and parallel to the line $x - 3y = 6$ meet the parabola ${y}^{2} = 4x$ at $B$,

So, the equation of line through $P$ will be $y - 3 = \dfrac{1}{3}\left(x - 1\right)$

$⇒ 3y = x + 8$

Now finding intersection of ${y}^{2} = 4x \text{ and } 3y = x + 8$ we get,

$B = \left(4, 4\right) \text{ and } \left(16, 8\right)$,

Also $B$ lies on the line $2x - 3y = 8,$ so \left(4, 4\right)$will not satisfy,</p><p>Hence,$B = \left(16, 8\right)$</p><p>So,${\left(AB\right)}^{2} = {\left(0 - 16\right)}^{2} + {\left(2 - 8\right)}^{2} = 256 + 36 = 292