Chapter 4.2, Problem 51E

To determine

To show: The curves $2x^2+3y^2=5\text{ and }{y}^2=x^3$ are orthogonal at $\left(1,1\right)\text{ and }\left(1,-1\right)$ . Draw the curves and show the tangent lines.

Answer to Problem 51E

The curves $2x^2+3y^2=5$and $y^2=x^3$are orthogonal.

Explanation of Solution

Given Information: Two curves $2x^2+3y^2=5\text{ and }{y}^2=x^3$ and two points $\left(1,1\right)\text{ and }\left(1,-1\right)$ .

Calculation:

The tangent to the curve $2x^2+3y^2=5$ is,

  $\begin{array}{c}2x^2+3y^2=5\\ 4x+6y\frac{dy}{dx}=0\\ \frac{dy}{dx}=-\frac{4x}{6y}\\ =-\frac{2x}{3y}\end{array}$

At point $\left(1,1\right)$ , the value of the derivative is,

  $\begin{array}{c}\frac{dy}{dx}=-\frac{2x}{3y}\\ =-\frac{2\left(1\right)}{3\left(1\right)}\\ =-\frac{2}{3}\end{array}$

The equation of the tangent line is,

  $\begin{array}{c}y-1=-\frac{2}{3}\left(x-1\right)\\ y-1=-\frac{2}{3}\left(x-1\right)\\ y=-\frac{2}{3}x+\frac{2}{3}+1\\ =-\frac{2}{3}x+\frac{5}{3}\end{array}$

At point $\left(1,-1\right)$ , the value of the derivative is,

  $\begin{array}{c}\frac{dy}{dx}=-\frac{2x}{3y}\\ =-\frac{2\left(1\right)}{3\left(-1\right)}\\ =\frac{2}{3}\end{array}$

The equation of the tangent line is,

  $\begin{array}{c}y-\left(-1\right)=\frac{2}{3}\left(x-1\right)\\ y+1=\frac{2}{3}\left(x-1\right)\\ y=\frac{2}{3}x-\frac{2}{3}-1\\ =\frac{2}{3}x-\frac{5}{3}\end{array}$

The tangent to the curve $y^2=x^3$ is,

  $\begin{array}{c}{y}^2=x^3\\ 2y\frac{dy}{dx}=3x^2\\ \frac{dy}{dx}=\frac{3x^2}{2y}\end{array}$

At point $\left(1,1\right)$ , the value of the derivative is,

  $\begin{array}{c}\frac{dy}{dx}=\frac{3x^2}{2y}\\ =\frac{3{\left(1\right)}^2}{2\left(1\right)}\\ =\frac{3}{2}\end{array}$

The equation of the tangent line is,

  $\begin{array}{c}y-1=\frac{3}{2}\left(x-1\right)\\ y-1=\frac{3}{2}\left(x-1\right)\\ y=\frac{3}{2}x-\frac{3}{2}+1\\ =\frac{3}{2}x-\frac{1}{2}\end{array}$

At point $\left(1,-1\right)$ , the value of the derivative is,

  $\begin{array}{c}\frac{dy}{dx}=\frac{3x^2}{2y}\\ =\frac{3{\left(1\right)}^2}{2\left(-1\right)}\\ =-\frac{3}{2}\end{array}$

The equation of the tangent line is,

  $\begin{array}{c}y-\left(-1\right)=-\frac{3}{2}\left(x-1\right)\\ y+1=-\frac{3}{2}\left(x-1\right)\\ y=-\frac{3}{2}x+\frac{3}{2}-1\\ =-\frac{3}{2}x+\frac{1}{2}\end{array}$

Two curves are orthogonal at a point of intersection if their tangents at that point cross at right angles.

Clearly, both the curves are orthogonal.

The curve along with the tangents are,

  

Red ellipse represents $2x^2+3y^2=5$ . Green and purple lines represent the tangent lines at the points $\left(1,1\right)\text{ and }\left(1,-1\right)$ respectively.

Light blue curve represents $y^2=x^3$ . Black and red lines represent the tangent lines at the points $\left(1,1\right)\text{ and }\left(1,-1\right)$ respectively.