Question

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

• A. $a+b$
• B. $2a+2b$
• C. $\sqrt{ab}$
• D. None of these

Answer 1

Answer: (A)

$a+b$

Let $x=asec\theta$ and $y=bcoec\theta$
Hence
$r^2=x^2+y^2$
$=a^{2}cose{c}^{2}t+b^{2}se{c}^{2}t$
$=a^2(1+co{t}^{2}t)+b^2(1+ta{n}^{2}t)$
$=(a^2+b^2)+(a^{2}co{t}^{2}t+b^{2}ta{n}^{2}t)$
$=(a+b{)}^2+(a^{2}co{t}^{2}t+b^{2}ta{n}^{2}t-2ab)$
$=(a+b{)}^2+(acot-btant{)}^2$
Hence minimum value will be $(a+b{)}^2$
Hence maximum value of $r_{min}^2=(a+b{)}^2$
Hence $r_{min}=a+b$.