A) \[\left( \frac{1}{2},\,\frac{15}{4} \right)\]
B) \[\left( -\frac{1}{2},\,\frac{15}{4} \right)\]
C) \[\left( \frac{1}{2},-\,\frac{15}{4} \right)\]
D) \[\left( -\frac{1}{2},-\,\frac{15}{2} \right)\]
Correct Answer: B
Solution :
Let the point of contact with ellipse is \[({{x}_{1}},\,\,{{y}_{1}})\]. Then, \[15x{{x}_{1}}+4y{{y}_{1}}-60=0\] ?(i) Let the tangent of \[{{y}^{2}}=8x\,\] is \[y=\,mx+\frac{2}{m}\] and the tangent of \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{15}=1\] is \[y=mx\pm \,\sqrt{4{{m}^{2}}+15}\] \[\therefore \] \[\frac{2}{m}=\pm \,\sqrt{4{{m}^{2}}+15}\] \[\Rightarrow \] \[4{{m}^{4}}+15{{m}^{2}}-4=0\] \[\Rightarrow \] \[m=\pm \frac{1}{2}\] Hence, equation of common tangent at \[m=\frac{1}{2}\] \[\Rightarrow \] \[y=\frac{x}{2}+4\] \[\Rightarrow \] \[x-2y+8=0\] ?(ii) From Eqs. (i) and (ii), we get \[\frac{15{{x}_{1}}}{1}=\frac{4{{y}_{1}}}{-2}=\frac{-60}{8}\] \[\Rightarrow \] \[{{x}_{1}}=-\frac{1}{2}\] and \[{{y}_{1}}=\frac{15}{4}\]
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