The equation of the tangent, to the curve $y=x^2-2x-3$ which is perpendicular to the line $x+2y+3=0$, is

$2x-y=7$

Line → $x+2y+3=0$

differentiating w.r.t x to find slope

$\frac{2dy}{dx}+1=0$ ⇒ slope $m$ = $\frac{dy}{dx}=\frac{-1}{2}$

slope tangent is n(tangent ⊥ line)

$⇒mn=-1⇒n×\frac{-1}{2}=-1$

$n=2$ ⇒ slope of tangent

curve → $y=x^2-2x-3$   ...(1)

$\frac{dy}{dx}=2x-2$ so $\frac{dy}{dx}=2⇒x=2$

from (1) when $x=2$

$y=-3$

at (2, -3) slope of tangent is 2

so equation of tangent: $(y-y_0)$ = slope $(x-x_0)$

$⇒y+3=2(x-2)$

$y+3=2x-4$

$2x-y=7$