Chapter 11.6, Problem 95AYU

To determine

To find: Equation of tangent line using Descartes’ method.

Answer to Problem 95AYU

Solution:

$y = - \frac{1}{3}x + \frac{7}{3}$

Explanation of Solution

Given:

$2x^2 + 3y^2 = 14$   –––(Eq 1)

And the point $(1,2)$

Calculation:

Equation of tangent line is $y = mx + b$

Using the fact that the point $(1,2)$ is on the line, we get

$\begin{array}{lll}b\hfill & = \hfill & 2 - m\hfill \\ y\hfill & = \hfill & mx + (2 - m)\hfill \end{array}$

Eq 1 implies, $2x^2 + 3y^2 = 14$

$\begin{array}{l}2x^2 + 3m^2{x}^2 + 12 + 3m^2 - 12m + 12mx - 6m^{2}x = 14\hfill \\ (2 + 3m^2)x^2 + (12m - 6m^2)x + 3m^2 - 12m - 2 = 0\hfill \end{array}$

Solving the above equation,

$x = \frac{ - (12m - 6m^2) \pm \sqrt{{(12m - 6m^2)}^2 - 4(2 + 3m^2)(3m^2 - 12m - 2)}}{2(2 + 3m^2)}$

Using Descartes’ method to obtain unique solution, discriminant $= 0$

$\begin{array}{rrr}\hfill {(12m - 6m^2)}^2 - 4(2 + 3m^2)(3m^2 - 12m - 2)& \hfill = & 0\hfill \\ \hfill 144m^2 + 96m + 16& \hfill = & 0\hfill \\ \hfill 9m^2 + 6m + 1& \hfill = & 0\hfill \\ \hfill m& = \hfill & \hfill - \frac{1}{3}\end{array}$

Equation of tangent line is $y = mx + b$

$m = - \frac{1}{3},b = 2 + \frac{1}{3} = \frac{7}{3}$

Therefore, $y = - \frac{1}{3}x + \frac{7}{3}$.