Question

A common tangent to $9x^2-16y^2=144and{x}^2+y^2=9$ is,

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

We are given a hyperbola ad a circle. we need to find the common tangent. We are going to use 2 aspects.

(1) Standard form of tangent to a hyperbola

(2) The distance between tangent to a circle and the centre of the circle is the radius of the circle.

The given equations are

$9x^2-16y^2=144$

i.e.,$\frac{x^2}{16}-\frac{y^2}{9}=1$ . . . . . . (Hyperbola)

$X^2+y^2=9$ . . . . . . (Circle)

lets draw the diagram

we known that the standard equation of a tangent of

slope 'm' to hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is

$y=mx\pm \sqrt{a^2{m}^2-b^2}$

for our case

$y=mx\pm \sqrt{16m^2-9}$ - - - - - (1)

since(1) is alc=so tangent to the circle $x^2+y^2=9$

$\left|\frac{0+0\pm \sqrt{16m^2-9}}{\sqrt{1+m^2}}\right|=3$

$\frac{16m^2-9}{1+m^2}=9$

$16m^2-9=9+9m^2$

$7m^2=18$

$m=\frac{3\sqrt{2}}{\sqrt{7}}$

$\therefore$ Required equation of tangent is,

$y=\frac{3\sqrt{2}}{\sqrt{7}}x\pm \sqrt{16.\frac{18}{7}-9}$

$y=3\sqrt{\frac{2}{7}}x\pm \sqrt{\frac{225}{7}}$

$=3\sqrt{\frac{2}{7}}x\pm \frac{15}{\sqrt{7}}$

hence option (b) is correct