Chapter 4, Problem 6P

Find the point on the parabola y = 1 − x² at which the tangent line cuts from the first quadrant the triangle with the smallest area.

To determine

To find: The points on the parabola $y=1-x^2$ at which the tangent line cuts from first quadrant the triangle with the smallest area.

Answer to Problem 6P

The point on the parabola is $\left(\frac{1}{\sqrt{3}},\frac{2}{3}\right)$

Explanation of Solution

Let the points on the curve be $\left(u,v\right)$.

As the points lie on the parabola $y=1-x^2$ write $v=1-u^2$.

Slope of the tangent line is, $\frac{dy}{dx}=-2x$.

Slope of the tangent line at point $\left(u,v\right)$ it is $-2u$.

Write the equation of the tangent line by using slope-point form.

Thus, the equation of the tangent line is, $y-v=-2u\left(x-u\right)$.

Find x-intercepts and y-intercepts of the tangent.

To find x-intercept, put $y=0$.

Thus, the x-intercept is $u+\frac{v}{2u}$.

Similarly to find y-intercept, put $x=0$.

Thus, the y-intercept is $v+2u^2$.

Now area of the triangle formed is given by,

$\begin{array}{c}A=\frac{1}{2}\cdot x-\text{intercept}\cdot y-\text{intercept}\\ \text{=}\frac{1}{2}\cdot \left(u+\frac{v}{2u}\right)\left(v+2u^2\right)\\ =\frac{{\left(2u^2+v\right)}^2}{4u}\end{array}$

Substitute value of v as $v=1-u^2$.

$\begin{array}{c}A=\frac{{\left(2u^2+v\right)}^2}{4u}\\ =\frac{{\left(2u^2+1-u^2\right)}^2}{4u}\\ =\frac{{\left(1+u^2\right)}^2}{4u}\end{array}$

Differentiate A with respect to u.

$\begin{array}{c}\frac{dA}{du}=\frac{d}{du}\left(\frac{{\left(1+u^2\right)}^2}{4u}\right)\\ =\frac{16u^2\left(1+u^2\right)-{\left(1+u^2\right)}^{2}4}{16u^2}\end{array}$

Set $\frac{dA}{du}=0$ to get,

$\begin{array}{l}\frac{16u^2\left(1+u^2\right)-{\left(1+u^2\right)}^{2}4}{16u^2}=0\\ 16u^2\left(1+u^2\right)-{\left(1+u^2\right)}^{2}4=0\\ 16u^2\left(1+u^2\right)={\left(1+u^2\right)}^{2}4\\ 16u^2=\left(1+u^2\right)4\end{array}$

Simplify further as follows.

$\begin{array}{l}16u^2=4\left(1+u^2\right)\\ 12u^2=4\\ u^2=\frac{4}{12}\\ u=\pm \frac{1}{\sqrt{3}}\end{array}$

As the point is in first quadrant $u=\frac{1}{\sqrt{3}}$.

Substitute to get value of v.

$\begin{array}{c}v=1-u^2\\ =1-{\left(\frac{1}{\sqrt{3}}\right)}^2\\ =1-\frac{1}{3}\\ =\frac{2}{3}\end{array}$

Hence, for the values $u=\frac{1}{\sqrt{3}}\text{ and v}=\frac{2}{3}$ the tangent line to the curve $y=1-x^2$ cuts the first quadrant in a triangle with smallest area.