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 $r$,
$\frac{d\left(r^{2}\right)}{d y} =0$
$\Rightarrow \frac{-2 y b^{2} a^{2}}{\left(y^{2}-b^{2}\right)^{2}}+2 y =0$
$\Rightarrow y^{2} =b(a +b)$
$\therefore x^{2} =a(a+ b)$
$\Rightarrow r^{2}=(a +b)^{2}$
$\Rightarrow r =a+ b$