Equation of the tangents to the ellipse $\frac{{\mathrm{x}}^2}{4}+\frac{{\mathrm{y}}^2}{3}=1$. perpendicular to the line 2x + y + 7 = 0 are :

slope of given line  2x+y+7=0 is -2

Slope of the line perpendicular to 2x+y+7=0 is 1/2

equation of  tangent to the ellipse $\frac{{\mathrm{x}}^2}{4}+\frac{{\mathrm{y}}^2}{3}=1$ having slope 1/2 is

$$\begin{gathered} y=mx\pm \sqrt{a^2{m}^2+b^2} \\ \Rightarrow y=\frac{1}{2}x\pm \sqrt{4\left(\frac{1}{4}\right)+3}\Rightarrow y=\frac{1}{2}x\pm 2 \\ \Rightarrow x-2y\pm 4=0 \end{gathered}$$