QJEEAdv10Ma5678: JEE Advanced 2010 Mathematics - Class 11 JEEAdv10Ma5678 Solution: The circle and hyperbola intersect at the points and . Equation …

A tangent to ${{{x^2}} \over 9} - {{{y^2}} \over 4} = 1$ is

y = mx + \sqrt {9{m^2} - 4} ,\,m > 0 .... (1)

A tangent to ${(x - 4)^2} + {y^2} = 16$ is

$xy = m(x - 4) + 4\sqrt {1 + {m^2}}$ ..... (2)

Comparing (1) and (2),

$\sqrt {9{m^2} - 4} = - 4m + 4\sqrt {1 + {m^2}} \Rightarrow \sqrt {9 - {4 \over {{m^2}}}} = - 4 + 4\sqrt {1 + {1 \over {{m^2}}}}$

Let ${1 \over {{m^2}}} = t$ , we have $\sqrt {9 - 4t} = - 4 + 4\sqrt {1 + t}$

Squaring, we have

$\Rightarrow 9 - 4t = 16 + 16(1 + t) - 32\sqrt {1 + t} \Rightarrow 32\sqrt {1 + t} = 23 + 20t$

Again squaring $1024(1 + t) = 529 + 920t + 400{t^2}$

$\Rightarrow 400{t^2} - 104t - 495 = 0 \Rightarrow t = {5 \over 4}$

Thus ${m^2} = {4 \over 5},\,m = {2 \over {\sqrt 5 }}$

The tangent is $y = {2 \over {\sqrt 5 }}x + {4 \over {\sqrt 5 }}$ i.e. $2x - \sqrt 5 y + 4 = 0$