A) \[y=\frac{3}{\sqrt{7}}x+\frac{15}{\sqrt{7}}\]
B) \[y=3\sqrt{\frac{2}{7}}x+\frac{15}{\sqrt{7}}\]
C) \[y=2\sqrt{\frac{3}{7}}x+15\sqrt{7}\]
D) \[y=\frac{3}{\sqrt{7}}x-\frac{15}{\sqrt{7}}\]
Correct Answer: B
Solution :
Given curves are \[9{{x}^{2}}-16{{y}^{2}}=144\] and \[{{x}^{2}}+{{y}^{2}}=9\] Let the equation of common tangent be \[y=mx+c\] Since \[y=mx+c\] is tangent to \[{{x}^{2}}+{{y}^{2}}=9,\] so \[{{c}^{2}}=9(1+{{m}^{2}})\] ?(1) Also, \[y=mx+c\]is tangent to \[\frac{{{x}^{2}}}{16}-\frac{{{y}^{2}}}{9}=1\] So, \[{{c}^{2}}=16{{m}^{2}}-9\] ?(2) \[\therefore \]From (1) and (2), we get \[16{{m}^{2}}-9=9+9{{m}^{2}}\] \[\Rightarrow {{m}^{2}}=\frac{18}{7}\Rightarrow m=\pm \sqrt{\frac{2}{7}}\] Also, \[c=\pm \frac{15}{\sqrt{7}}\] Hence, \[y=3\sqrt{\frac{2}{7}x}+\frac{15}{\sqrt{7}}\] is one of their common tangent.
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