• # question_answer The circle ${{x}^{2}}+{{y}^{2}}-8x=0$ and hyperbola $\frac{{{x}^{2}}}{9}-\frac{{{y}^{2}}}{4}=1.$ The equation of a common tangent with positive slope to the circle as well as to the hyperbola is A) $2x-\sqrt{5}y-20=0$      B) $2x-\sqrt{5}y+4=0$ C) $3x-4y+8=0$   D) $4x-3y+4=0$

Solution :

[b]  Equation of tangent $\frac{{{x}^{2}}}{9}-\frac{{{y}^{2}}}{4}=1$ $y=mx+\sqrt{9{{m}^{2}}-4},m>0$ Equation of circle ${{(x-4)}^{2}}+{{y}^{2}}=16$is $y=mx-4m+4\sqrt{1+{{m}^{2}}}$ Equating the constant terms, we get $\sqrt{9{{m}^{2}}-4}=-4m+4\sqrt{1+{{m}^{2}}}$ On squaring, we get $9{{m}^{2}}-4=16{{m}^{2}}-32m+\sqrt{1+{{m}^{2}}}+16+16\,{{m}^{2}}$ $=23{{m}^{2}}+20=32\,m\sqrt{1+{{m}^{2}}}$
 On squaring again, we get $529{{m}^{4}}+920{{m}^{2}}+400=1024{{m}^{2}}+1024{{m}^{4}}$ $\Rightarrow$$495{{m}^{4}}+104{{m}^{2}}-400=0$ $\Rightarrow$$(5{{m}^{2}}-4)(99{{m}^{2}}+100)=0$ $\Rightarrow$${{m}^{2}}=\frac{4}{5}$ $\Rightarrow$$m=\frac{2}{\sqrt{5}}$ Equation of tangent $=y=\frac{2}{\sqrt{5}}x+\frac{4}{\sqrt{5}}$ $\Rightarrow$$2x-\sqrt{5}y+4=0$

You need to login to perform this action.
You will be redirected in 3 sec