A) \[x+y=1\]
B) \[x-y=1\]
C) \[x+y=\frac{9}{2}\]
D) \[x-y=\frac{3}{2}\]
Correct Answer: D
Solution :
The equation of tangent of slope m to \[{{x}^{2}}=6y,\] is \[y=mx-\frac{3}{2}{{m}^{2}}\] Now, using condition of tangency, we get \[\frac{9}{4}{{m}^{4}}=\frac{9}{2}{{m}^{2}}-\frac{9}{4}\Rightarrow \,{{m}^{4}}=2{{m}^{2}}-1\] \[\Rightarrow \,{{m}^{4}}-2{{m}^{2}}+1=0\] \[\Rightarrow \,{{({{m}^{2}}-1)}^{2}}=0\Rightarrow m=\,\underline{+}\,1\] \[\therefore \]For m = 1, we get equation of common tangent as\[x-y=\frac{3}{2}\].
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