Question

Shortest distance between the curves $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,4x^2+4y^2=a^2(b>a)$ is

• A. $\frac{b}{2}$
• B. $\frac{b}{\sqrt{2}}$
• C. $\frac{a}{2}$
• D. $\frac{a}{\sqrt{2}}$

Answer 1

The correct option is C $\frac{a}{2}$

The curve $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is a hyperbola with center $(0,0)$, the transverse and conjugate axes of hyperbola are same as coordinate axes.

The given circle $4x^2+4y^2=a^2$ has also it's center on $(0,0)$. The radius of circle is $\frac{a}{2}$,

The cicle cuts the $x-$axis at $(\frac{a}{2},0)$,

From the figure we can see the shortest diatnce between the hyperbola and circle is $a-\frac{a}{2}=\frac{a}{2}$