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. $a+b$
• C. $2a+b$
• D. None of these

Answer 1

The correct option is B $a+b$

Given curve is $\frac{a^2}{x^2}+\frac{b^2}{y^2}=1$
Let radius vector is ${}^{\prime }{r}^{\prime }$
Therefore, $r^2=x^2+y^2$
$\Rightarrow r^2=\frac{a^2{y}^2}{y^2-b^2}+y^2(\because \frac{a^2}{x^2}+\frac{b^2}{y^2}=1)$
For minimum value of $r$,
$\frac{d\left(r^2\right)}{dy}=0$
$\Rightarrow \frac{-2y{b}^2{a}^2}{(y^2-b^2{)}^2}+2y=0$
$\Rightarrow y^2=b(a+b)$
Thus $x^2=a(a+b)$
$\Rightarrow r^2=(a+b{)}^2$
$\Rightarrow r=a+b$.