Question

The minimum radius vector of the curve $\frac{a^2}{x^2}+\frac{b^2}{y^2}=1$ is of length

• A. $a-b$
• B. $\left|a+b\right|$
• C. $2a+b$
• D. None of these

Answer 1

The correct option is B

$\left|a+b\right|$

Explanation for the correct option:

Step 1: Find the critical points.

An equation of curve $\frac{a^2}{x^2}+\frac{b^2}{y^2}=1$ is given.

Assume that, $x=r\cos \theta$ and $y=r\sin \theta$.

Therefore, the given equation becomes:

$$\begin{gathered} \frac{a^2}{{\left(r\cos \theta \right)}^2}+\frac{b^2}{{\left(r\sin \theta \right)}^2}=1 \\ \Rightarrow \frac{a^2}{r^2{\cos }^2\theta }+\frac{b^2}{r^2{\sin }^2\theta }=1 \\ \Rightarrow r^2=\frac{a^2}{{\cos }^2\theta }+\frac{b^2}{{\sin }^2\theta } \\ \Rightarrow r^2=a^{2}se{c}^2\theta +b^2\cos e{c}^2\theta \end{gathered}$$

Differentiate both sides with respect to $\theta$.

$\frac{d{r}^2}{d\theta }=2a^{2}sec\theta ·sec\theta ·\tan \theta -2b^2\cos ec\theta ·\cos ec\theta ·cot\theta$

Substitute $\frac{d{r}^2}{d\theta }=0$ to find the critical points.

$$\begin{gathered} 2a^{2}se{c}^2\theta ·\tan \theta -2b^2\cos e{c}^2\theta ·cot\theta =0 \\ \Rightarrow a^{2}se{c}^2\theta ·\tan \theta =b^2\cos e{c}^2\theta ·cot\theta \\ \Rightarrow a^2\left(\frac{1}{{\cos }^2\theta }\right)·\frac{\sin \theta }{\cos \theta }=b^2\frac{1}{{\sin }^2\theta }·\frac{\cos \theta }{\sin \theta } \\ \Rightarrow a^2\frac{\sin \theta }{{\cos }^3\theta }=b^2·\frac{\cos \theta }{{\sin }^3\theta } \\ \Rightarrow \frac{{\sin }^4\theta }{{\cos }^4\theta }=\frac{b^2}{a^2} \\ \Rightarrow {\tan }^2\theta =\frac{b}{a} \end{gathered}$$

Therefore, ${\tan }^2\theta =\frac{b}{a}$ is the critical point.

Step 2: Find the minimum radius vector.

Evaluate $r$ for ${\tan }^2\theta =\frac{b}{a}$.

Since,

$$\begin{gathered} r^2=a^{2}se{c}^2\theta +b^2\cos e{c}^2\theta \\ \Rightarrow r^2=a^2\left(1+{\tan }^2\theta \right)+b^2\left(1+co{t}^2\theta \right) \end{gathered}$$

Since, ${\tan }^2\theta =\frac{b}{a}$.

$$\begin{gathered} \Rightarrow {r^2}_{min}=a^2\left(1+\frac{b}{a}\right)+b^2\left(1+\frac{a}{b}\right) \\ \Rightarrow {r^2}_{min}=a^2\left(\frac{a+b}{a}\right)+b^2\left(\frac{a+b}{b}\right) \\ \Rightarrow {r^2}_{min}=a\left(a+b\right)+b\left(a+b\right) \\ \Rightarrow {r^2}_{min}=\left(a+b\right)\left(a+b\right) \\ \Rightarrow r_{min}=\left|a+b\right| \end{gathered}$$

Therefore, the minimum radius vector of the given curve is $\left|a+b\right|$.

Hence, option $B$ is the correct answer.