A) \[2{{y}^{2}}=9x\]
B) \[{{y}^{2}}=9x\]
C) \[2{{y}^{2}}=27x\]
D) \[{{y}^{2}}=15x\]
Correct Answer: C
Solution :
Let \[P(h,\,k)\] be the point from which two tangent are drawn to\[{{y}^{2}}=12x\]Any tangent by\[{{y}^{2}}=12x\]is \[T:y=mx+\frac{3}{m}\] \[{{m}^{2}}h=mk+3=0\] \[{{m}_{1}}+{{m}_{2}}=\frac{k}{h};{{m}_{1}}{{m}_{2}}=\frac{3}{h}\]\[{{m}_{1}}=2{{m}_{2}}\](given) \[\therefore \,\,3{{m}_{3}}=\frac{k}{h}\Rightarrow \,{{m}_{3}}=\frac{k}{3h}\] \[2m_{2}^{2}=\frac{3}{h}\Rightarrow \,2.\frac{{{k}^{2}}}{9{{h}^{2}}}=\frac{3}{h}\] \[\Rightarrow \]\[{{y}^{2}}=\frac{27}{2}x\]Ans.
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