Use implicit differentiation to find an equation of the tangent line to the ellipse $\frac{x^{2}}{2}+\frac{y^{2}}{242}=1$ at $(1,11)$ $y=-8 x+15$ $y=-12 x+13$ $y=-6 x+13$ $y=-7 x+15$ $y=-11 x+22$

Step 1 ：First, we need to find the derivative of the given equation. The derivative of \(\frac{x^{2}}{2}+\frac{y^{2}}{242}=1\) with respect to \(x\) is obtained using implicit differentiation. The derivative of \(\frac{x^{2}}{2}\) with respect to \(x\) is \(x\), and the derivative of \(\frac{y^{2}}{242}\) with respect to \(x\) is \(\frac{y}{121}y'\), where \(y'\) is the derivative of \(y\) with respect to \(x\). So, the derivative of the given equation is \(x+\frac{y}{121}y'=0\).

Step 2 ：Next, we solve the derivative equation for \(y'\). We get \(y'=-\frac{121x}{y}\).

Step 3 ：Then, we substitute the point \((1,11)\) into the derivative equation to find the slope of the tangent line at this point. We get \(y'=-\frac{121*1}{11}=-11\).

Step 4 ：So, the slope of the tangent line at the point \((1,11)\) is \(-11\).

Step 5 ：Finally, we use the point-slope form of the equation of a line to find the equation of the tangent line. The point-slope form is \(y-y_{1}=m(x-x_{1})\), where \(m\) is the slope and \((x_{1},y_{1})\) is a point on the line. Substituting \(m=-11\) and \((x_{1},y_{1})=(1,11)\), we get \(y-11=-11(x-1)\), which simplifies to \(y=-11x+22\).

Step 6 ：So, the equation of the tangent line to the ellipse at the point \((1,11)\) is \(\boxed{y=-11x+22}\).