Question

(a) Use implicit differentiation to find an equation of the tangent line to the ellipse \(\frac{x^{2}}{2}+\frac{y^{2}}{8}=1\) at (1,2) (b) Show that the equation of the tangent line to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) at \(\left(x_{0}, y_{0}\right)\) is \(\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}=1\)

Short answer

(a) The equation of the tangent line to the ellipse \(\frac{x^{2}}{2}+\frac{y^{2}}{8}=1\) at (1,2) is \(4x-2y=2\). (b) The equation of the tangent line to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) at \(\left(x_{0}, y_{0}\right)\) is indeed \(\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}=1\) as shown in step 6.

Step-by-step solution

Implicit Differentiation for the first Equation

Begin by differentiating both sides of the ellipse equation \(\frac{x^{2}}{2}+\frac{y^{2}}{8}=1\) with respect to x. Use implicit differentiation: differentiate \(\frac{x^{2}}{2}\) to get \(x\) and differentiate \(\frac{y^{2}}{8}\) with respect to x to get \(\frac{y}{4}*y'\), where \(y'\) is the derivative of y with respect to x. Set the result equal to zero, since the derivative of a constant is always zero.

Solve for Derivative

Next, solve the resulting equation from Step 1 for \(y'\), which represents the derivative of y with respect to x. \(y'\) will be the slope of the tangent line at a given point.

Calculate Slope of Tangent at Given Point (1,2)

When \(y'\) has been determined, substitute the coordinates of the given point (1,2) into the equation to find the slope of the tangent line at that point.

Construct Equation of Tangent

The equation of a straight line is given by \(y=mx+c\), where m is the slope, and c is the intercept. Substitute the slope found in the previous step and the coordinates (1,2) into this equation and solve it to find the equation of the tangent line.

Implicit Differentiation for the Second Equation

Differentiate both sides of the general ellipse equation \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) with respect to x, likewise to Step 1. Solve the resulting equation for \(y'\) again.

Show Equation of the Tangent Line

Substitute \(y'\), \(x_{0}\), and \(y_{0}\) into the formula \(y-y_{0}=m(x-x_{0})\) for the equation of a line, to show that the equation of the tangent line to the ellipse at \(\left(x_{0}, y_{0}\right)\) is \(\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}=1\).