Question

The length of the line segment when tangent of the curve intersect to both axes for curve ${\left(\frac{x}{a}\right)}^{\frac{2}{3}}={\cos }^2\left(\theta \right)$ and ${\left(\frac{y}{a}\right)}^{\frac{2}{3}}={\sin }^2\left(\theta \right)$ is

• A. $a$
• B. $\left|a\right|$
• C. $a^2$
• D. $a^3$

Answer 1

The correct option is A

$a$

Explanation for correct option:

Step 1. Find the slope

Given that ${\left(\frac{x}{a}\right)}^{\frac{2}{3}}={\cos }^2\left(\theta \right)$ and ${\left(\frac{y}{a}\right)}^{\frac{2}{3}}={\sin }^2\left(\theta \right)$

$$\begin{gathered} \therefore {\left(\frac{x}{a}\right)}^{\frac{2}{3}}={\cos }^2\left(\theta \right) \\ \Rightarrow {\left(\frac{x}{a}\right)}^{\frac{1}{3}}=\cos \left(\theta \right) \\ \Rightarrow x=a{\cos }^3\left(\theta \right)..........\left(i\right) \\ \Rightarrow \frac{dx}{d\theta }=-3a{\cos }^2\left(\theta \right)\sin \left(\theta \right) \\ {\left(\frac{y}{a}\right)}^{\frac{2}{3}}={\sin }^2\left(\theta \right) \\ \Rightarrow {\left(\frac{y}{a}\right)}^{\frac{1}{3}}=\sin \left(\theta \right) \\ \Rightarrow y=a{\sin }^3\left(\theta \right)......\left(ii\right) \\ \Rightarrow \frac{dy}{d\theta }=3a{\sin }^2\left(\theta \right)\cos \left(\theta \right) \\ \therefore \frac{dy}{dx}=-\tan \left(\theta \right) \end{gathered}$$

Step 2. Find the equation of the tangent

From $\left(i\right)$ and $\left(ii\right)$ we get $\left(x_1,y_1\right)=\left(a{\cos }^3\left(\theta \right),a{\sin }^3\left(\theta \right)\right)$

Now the Equation of the tangent at point $\left(x_1,y_1\right)$ is

$$\begin{gathered} \left(y-y_1\right)=\frac{dy}{dx}\left(x-x_1\right) \\ \Rightarrow \left(y-a{\sin }^3\left(\theta \right)\right)=-\tan \left(\theta \right)\left(x-a{\cos }^3\left(\theta \right)\right) \\ \Rightarrow \left(y-a{\sin }^3\left(\theta \right)\right)=-\frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}\left(x-a{\cos }^3\left(\theta \right)\right) \\ \Rightarrow y\cos \left(\theta \right)-a\cos \left(\theta \right){\sin }^3\left(\theta \right)=-x\sin \left(\theta \right)+a\sin \left(\theta \right){\cos }^3\left(\theta \right) \\ \Rightarrow y\cos \left(\theta \right)+x\sin \left(\theta \right)=a\cos \left(\theta \right)\sin \left(\theta \right)\left({\sin }^2\left(\theta \right)+{\cos }^2\left(\theta \right)\right) \\ \Rightarrow y\cos \left(\theta \right)+x\sin \left(\theta \right)=a\cos \left(\theta \right)\sin \left(\theta \right) \\ \Rightarrow \frac{y}{\sin \left(\theta \right)}+\frac{x}{\cos \left(\theta \right)}=a \end{gathered}$$

Step 3. Find the length of tangent

The equation of the tangent is $\frac{y}{a\sin \left(\theta \right)}+\frac{x}{a\cos \left(\theta \right)}=1$

Therefore the tangent intercept the $x$axis at $A\left(a\cos \left(\theta \right),0\right)$ and the $y$ axis at $B\left(0,a\sin \left(\theta \right)\right)$

$\begin{array}{rcl}\therefore AB& =& \sqrt{{\left(a\cos \left(\theta \right)\right)}^2+{\left(a\sin \left(\theta \right)\right)}^2}\\ & =& \sqrt{a^2\left({\cos }^2\left(\theta \right)+{\sin }^2\left(\theta \right)\right)}\\ & =& \sqrt{a^2}\\ & =& a\end{array}$

Hence, the correct option is option A.