Let the tangent to the curve ${\mathrm{x}}^2+2\mathrm{x}-4\mathrm{y}+9=0$ at the point P(1,3) on it meet the y-axis at A. Let the line passing through P and parallel to the line x – 3y = 6 meet the parabola ${\mathrm{y}}^2=4\mathrm{x}$ at B. If B lies on the line 2x – 3y = 8, then $(\mathrm{AB}{)}^2$ is equal to _____

$4\mathrm{y}={\mathrm{x}}^2+2\mathrm{x}+9$

$4\mathrm{y}-8=(\mathrm{x}+1{)}^2$

$4(\mathrm{y}-2)=(\mathrm{x}+1{)}^2\mathrm{a}=1$

Tangent is  $(\mathrm{y}-2)=\mathrm{m}(\mathrm{x}+1)-{\mathrm{am}}^2$

$\mathrm{y}-2=\mathrm{m}(\mathrm{x}+1)-{\mathrm{m}}^2$

Passing through P(1, 3)

$3-2=\mathrm{m}(1+1)-{\mathrm{m}}^2$

$(\mathrm{m}-1{)}^2=0$

$m = 1$

$\mathrm{y}-2=1\left(\mathrm{x}+1\right)-1$

$x-y+2=0$

Intersects y-axis at A (0, 2)

B lies on 2x – 3y = 8

Equation of line parallel to x – 3y = 6

$\text{x-3y+k=0}$

sub P(1,3)

k=8

$\mathrm{x}-3\mathrm{y}+8=0$

Point B(x, y) is intersection of lines 2x – 3y = 8 and x-3y+8=0

We set B (16, 8)

$(\mathrm{AB}{)}^2=(16-0{)}^2+(8-2{)}^2=292$