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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2001

  • question_answer
    The equation of tangent at \[(-4,-4)\] on the curve\[{{x}^{2}}=-4y\] is:

    A) \[2x+y+4=0\]    

    B) \[2x-y-12=0\]

    C) \[2x+y-4=0\]     

    D) \[2x-y+4=0\]

    Correct Answer: D

    Solution :

    The equation of curve \[{{x}^{2}}=-4y.\] Differentiating the given equation \[2x=-4\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=-\frac{x}{2}=\frac{4}{2}=2\]the equation of tangent \[y-{{y}_{1}}=\frac{dy}{dx}(x-{{x}_{1}})\] \[y+4=2(x+4)\Rightarrow y+4=2x+8\] \[\Rightarrow \]               \[2x-y+4=0\]


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