Question

The arrangement of the following hyperbolas in the ascending order of the radius of their director circles.
A: $16x^2-25y^2=400$
B: $4x^2-9y^2=36$
C: $9x^2-16y^2=144$
D: $3x^2-4y^2=12$

• A. $A,B,C,D$
• B. $A,C,B,D$
• C. $D,B,C,A$
• D. $D,C,B,A$

Answer 1

The correct option is C $D,B,C,A$

We know that a Director Circle is the locus of point of intersection of a pair of perpendicular tangents to a hyperbola.

So if the center of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is at origin, then the Director circle is a circle centered at origin and radius $r=\sqrt{a^2-b^2}$

So we find radius for all given hyperbolas.

A. ${16x}^2-{25y}^2=400\Rightarrow \frac{x^2}{25}-\frac{y^2}{16}=1$

So radius $r=\sqrt{{\left(5\right)}^2-{\left(4\right)}^2}=3$

B. ${4x}^2-{9y}^2=36\Rightarrow \frac{x^2}{9}-\frac{y^2}{4}=1$

So radius $r=\sqrt{{\left(3\right)}^2-{\left(2\right)}^2}=\sqrt{5}$

C. ${9x}^2-{16y}^2=144\Rightarrow \frac{x^2}{16}-\frac{y^2}{9}=1$

So radius $r=\sqrt{{\left(4\right)}^2-{\left(3\right)}^2}=\sqrt{7}$

D. ${3x}^2-{4y}^2=12\Rightarrow \frac{x^2}{4}-\frac{y^2}{3}=1$

So radius $r=\sqrt{{\left(2\right)}^2-{\left(\sqrt{3}\right)}^2}=1$

So ascending order or radius is DBCA.