Find the equations of the tangent and normal to the ellipse 2x² + 3y²  = 11 at the point whose ordinate is “1”

Equation of the ellipse is $2{\mathrm{x}}^2+3{\mathrm{y}}^2=11$
$\text{ Given ordinate }\mathrm{y}=1then 2{\mathrm{x}}^2+3.1=11\Rightarrow 2{\mathrm{x}}^2=8$
${\mathrm{x}}^2=4\Rightarrow {\mathrm{x}}^2=\pm 2$
points on the ellipse are P(2,1) and Q(–2,1)
Case (1) : P(2,1)
Equation of the tangent is
$$\begin{gathered} 2\mathrm{x}\cdot 2+3\mathrm{y}\cdot 1=11\Rightarrow 4\mathrm{x}+3\mathrm{y}=11 \\ \Rightarrow 14\mathrm{x}+3\mathrm{y}-11=0 \end{gathered}$$
since normal is $\perp$lr to the tangent then  equation of the normal at P can be taken as 3x - 4y = K
normal passes through P(2,1) then
$6-4=\mathrm{K}\Rightarrow \mathrm{K}=2$
equation of the normal at P is $3\mathrm{x}-4\mathrm{y}=2$
Case (2) : Q(–2, 1)
equation of the tangent at Q is
$\begin{array}{l}2\mathrm{x}(-2)+3\mathrm{y}\cdot 1=11\\ \Rightarrow -4\mathrm{x}+3\mathrm{y}=11\\ \Rightarrow 4\mathrm{x}-3\mathrm{y}+11=0\end{array}$
equation of the normal can be taken as
$3\mathrm{x}+4\mathrm{y}=\mathrm{K}$
The normal passes through Q(–2,1)
$-6+4=\mathrm{K}\Rightarrow \mathrm{K}=-2$
equation of the normal at Q is
$3\mathrm{x}+4\mathrm{y}=-2\text{ or }3\mathrm{x}+4\mathrm{y}+2=0$