# Consider the curve x^{2/3} + y^{2/3} = 4. Let L be the tangent line to this curve at the point...

## Question:

Consider the curve {eq}\displaystyle\; x^{2/3} + y^{2/3} = 4 {/eq}.

Let {eq}L\, {/eq} be the tangent line to this curve at the point {eq}(1,3\sqrt{3}) {/eq}, and let {eq}A\, {/eq} and {eq}B\, {/eq} be the {eq}x {/eq}- and {eq}y {/eq}-intercepts of {eq}L {/eq}.

What is the length of the line segment {eq}AB {/eq}?

(a) {eq}\displaystyle\;8 {/eq}

(b) {eq}\displaystyle\;\frac{\sqrt{3}}{8} {/eq}

(c) {eq}\displaystyle\;\frac{8}{\sqrt{3}} {/eq}

(d) {eq}\displaystyle\;\sqrt{3} {/eq}

(e) {eq}\displaystyle\;\frac{3^{1/3}}{8} {/eq}

## Equation of a Tangent Line, Intercepts of the Line and the Distance between Two Points :

The equation of a line which passes through the point {eq}\left (x_{1}, y_{1} \right ) {/eq} and having the slope {eq}m {/eq} is given by:

{eq}y-y_{1}=m (x-x_{1}) {/eq}

Slope {eq}m {/eq} is equal to the value of the derivative of the function at the given point.

For the x-intercept we put {eq}y=0 {/eq} and for the y-intercept we put {eq}x=0 {/eq} in the equation of the line.

The distance between two points can be found by using the distance formula.

The given curve is:

{eq}x^{2/3} + y^{2/3} = 4 {/eq}

Upon differentiating the above equation with respect to {eq}x {/eq} we get:

{eq}\\\\\begin{align*} \frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3}\frac{dy}{dx} & = 0 \\\\\Rightarrow x^{-1/3} + y^{-1/3}\frac{dy}{dx} & = 0 \\\\\Rightarrow \frac{1}{x^{1/3}}+\frac{1}{y^{1/3}}\frac{dy}{dx} & = 0 \\\\\Rightarrow \frac{1}{y^{1/3}}\frac{dy}{dx} & = -\frac{1}{x^{1/3}} \\\\\Rightarrow \frac{dy}{dx} & = -\frac{y^{1/3}}{x^{1/3}} \\\\\Rightarrow \frac{dy}{dx} & = -\left(\frac{y}{x}\right )^{1/3} \\\\\end{align*} {/eq}

Thus, the slope {eq}m {/eq} of the tangent line to the given curve at the point {eq}(1,3\sqrt{3}) {/eq} is:

{eq}\\\\\begin{align*} m & = -\left(\frac{3\sqrt{3}}{1}\right )^{1/3} \\\\& = -\left(3^{3/2}\right )^{1/3} \\\\& = -(3)^{1/2} \\\\& = - \sqrt{3} \\\\\end{align*} {/eq}

Thus, the equation of the tangent line which passes through the point {eq}(1,3\sqrt{3}) {/eq} and having the slope as {eq}- \sqrt{3} {/eq} is :

{eq}\\\\\begin{align*} y-3\sqrt{3} & = - \sqrt{3} (x-1) \\\\\Rightarrow y & = - \sqrt{3} x + \sqrt{3} +3\sqrt{3} \\\\\Rightarrow y & = - \sqrt{3} x +4\sqrt{3} \\\\\end{align*} {/eq}

Thus, the x and the y intercepts of the line are:

{eq}A\left (4, 0 \right ), \ \ B\left ( 0, 4\sqrt{3} \right ) {/eq}

Thus, the length of {eq}AB {/eq} is:

{eq}\\\\\begin{align*} AB & = \sqrt{4^{2}+(4\sqrt{3})^{2}} & \texttt{(Using the distance formula)} \\\\& = \sqrt{16+48} \\\\& = \sqrt{64} \\\\& = 8 \\\\\end{align*} {/eq}

Hence, option {eq}(a) {/eq} is correct. 