Q45 / 84 · JEE Advanced 2010 (Paper 1) · Mathematics
Paragraph:
The circle $x^2+y^2-8 x=0$ and hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ intersect at the points $A$ and $B$.Question:
Equation of a common tangent with positive slope to the circle as well as to the hyperbola is
- A$2 x-\sqrt{5} y-20=0$
- Correct answerB$2 x-\sqrt{5} y+4=0$
- C$3 x-4 y+8=0$
- D$4 x-3 y+4=0$
Solution
Equation of tangent to hyperbola having slope $m$ is
$$
y=m x+\sqrt{9 m^2-4}
$$
Equation of tangent to circle is
$$
y=m(x-4)+\sqrt{16 m^2+16}
$$
Eqs. (i) and (ii) will be identical for $m=\frac{2}{\sqrt{5}}$ satisfy.
$\therefore$ Equation of common tangent is $2 x-\sqrt{5} y+4=0$
Hence, the correct answer is option: 'B'.