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. 2 a+ b
• D. None of these

Answer 1

Answer: (B)

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