JEE Main & Advanced Sample Paper JEE Main Sample Paper-32

  • question_answer
    A common tangent to the conies \[{{x}^{2}}=6y\] and \[2{{x}^{2}}-4{{y}^{2}}=9,\],is

    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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