A) \[x+y=1\]
B) \[x-y=1\]
C) \[x+y=\frac{9}{2}\]
D) \[x-y=\frac{3}{2}\]
Correct Answer: D
Solution :
Let \[y=(mx+c)\] is tangent to\[{{x}^{4}}=6y\] So, \[{{x}^{2}}-6mx-6c=0\] Put disc = 0 \[\Rightarrow c=\frac{-3}{2}{{m}^{2}}\] \[\therefore \] we get\[y=mx-\frac{3}{2}{{m}^{2}}\] Also, \[\frac{{{x}^{2}}}{9/2}-\frac{{{y}^{2}}}{9/4}=1\] So, using condition of tangency, we get \[\frac{9}{4}{{m}^{4}}=\frac{9}{2}{{m}^{2}}-\frac{9}{4}\] \[\Rightarrow \]\[{{m}^{4}}=2{{m}^{2}}-1\Rightarrow {{m}^{4}}-2{{m}^{2}}+1=0\] \[{{({{m}^{2}}-1)}^{2}}=0\Rightarrow m=\pm 1\] \[\therefore \]For\[m=1\], we get\[x-y=\frac{3}{2}\]
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