Chapter 3, Problem 47RE

(a).

To determine

To find the equation for the tangent.

Answer to Problem 47RE

The equation of tangent:

  $y_{\text{tangent}}=\frac{-3}{4\sqrt{2}}x+2+\frac{3}{4\sqrt{2}}$

Explanation of Solution

Given:

  $x^2+2y^2=9,\left(1,2\right)$

Concept Used:

tangent $=\frac{dy}{dx}$

Equation of tangent:

  $\begin{array}{l}y=mx+c,\\ m=\text{slope, c is constant}.\end{array}$

Calculation:

The values of $\left(x,y\right)$ :

  $\left(x,y\right)=\left(1,2\right)$

Rearrange the given equation:

  $\begin{array}{c}{x}^2+2y^2=9\\ y^2=\frac{9}{2}-\frac{x^2}{2}\\ y=\sqrt{\frac{9}{2}-\frac{x^2}{2}}\end{array}$

Slope of tangent is as follows:

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left[\sqrt{\frac{9}{2}-\frac{x^2}{2}}\right]\\ =\frac{1}{2}{\left(\frac{9}{2}-\frac{x^2}{2}\right)}^{\frac{1}{2}-1}\cdot \frac{d}{dx}\left[\frac{9}{2}-\frac{x^2}{2}\right]\\ =\frac{\frac{d}{dx}\left[\frac{9}{2}\right]-\frac{d}{dx}\left[\frac{x^2}{2}\right]}{2\sqrt{\left(\frac{9}{2}-\frac{x^2}{2}\right)}}\\ =\frac{-x}{2\sqrt{\left(\frac{9}{2}-\frac{x^2}{2}\right)}}\end{array}$

  $\begin{array}{c}{\left[\frac{dy}{dx}\right]}_{x=1}={\left[-\frac{x}{2\sqrt{\frac{9}{2}-\frac{x^2}{2}}}\right]}_{x=1}\\ =\frac{-1}{\frac{4\sqrt{2}}{3}}\\ =\frac{-3}{4\sqrt{2}}\end{array}$

The equation of tangent:

  $\begin{array}{c}{y}_{\text{tangent}}=2\\ \frac{-3}{4\sqrt{2}}x+c=2\\ \text{At}x=1:\\ \frac{-3}{4\sqrt{2}}\cdot 1+c=2\\ c=2+\frac{3}{4\sqrt{2}}\\ \text{Put}c=2+\frac{3}{4\sqrt{2}}\text{in}{y}_{\text{tangent}}\\ y_{\text{tangent}}=\frac{-3}{4\sqrt{2}}x+2+\frac{3}{4\sqrt{2}}\end{array}$

Conclusion:

The equation of tangent is:

  $y_{\text{tangent}}=\frac{-3}{4\sqrt{2}}x+2+\frac{3}{4\sqrt{2}}$

(b).

To determine

To find the equation of normal.

Answer to Problem 47RE

The equation of normal:

  $y_{normal}=\frac{4\sqrt{2}}{3}\left(x-1\right)+2$

Explanation of Solution

Given:

  $x^2+2y^2=9,\left(1,2\right)$

Concept Used:

The normal is perpendicular to the tangent.

Slope of normal:

  $=\frac{-1}{m}$

Calculation:

Slope of normal:

  $\begin{array}{l}=\frac{-1}{m}\\ =\frac{-1}{\frac{-3}{4\sqrt{2}}}\\ =\frac{4\sqrt{2}}{3}\end{array}$

  $m_{\text{normal}}=\frac{4\sqrt{2}}{3}$

Equation of normal:

  $\begin{array}{c}{y}_{normal}=\frac{4\sqrt{2}}{3}x+c\\ 2=\frac{4\sqrt{2}}{3}\cdot 1+c\\ c=2-\frac{4\sqrt{2}}{3}\\ y_{normal}=\frac{4\sqrt{2}}{3}x+2-\frac{4\sqrt{2}}{3}\\ y_{normal}=\frac{4\sqrt{2}}{3}\left(x-1\right)+2\end{array}$

Conclusion:

The equation of normal:

  $y_{normal}=\frac{4\sqrt{2}}{3}\left(x-1\right)+2$