A) \[2x-\sqrt{5}y-20=0\]
B) \[2x-\sqrt{5}y+4=0\]
C) \[3x-4y+8=0\]
D) \[4x-3y+4=0\]
Correct Answer: B
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 \[=y=\frac{2}{\sqrt{5}}x+\frac{4}{\sqrt{5}}\] \[\Rightarrow \]\[2x-\sqrt{5}y+4=0\] |
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