JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Self Evaluation Test - Application of Derivatives

  • question_answer
    The straight line \[\frac{x}{a}+\frac{y}{b}=2\] touches the curve \[{{\left( \frac{x}{a} \right)}^{n}}+{{\left( \frac{y}{b} \right)}^{n}}=2\] at the point (a, b) for

    A) n = 1, 2

    B) n = 3, 4, -5

    C) n = 1, 2, 3

    D) Any value of n

    Correct Answer: D

    Solution :

    [d] The point (a, b) lies on both the straight line and the given curve \[{{\left( \frac{x}{a} \right)}^{n}}+{{\left( \frac{y}{b} \right)}^{n}}=2\]. Differentiating the equation, we get \[\frac{dy}{dx}=-\frac{{{x}^{n-1}}}{{{a}^{n}}}.\frac{{{b}^{n}}}{{{y}^{n-1}}}\] \[\therefore {{\left( \frac{dy}{dx} \right)}_{at\,\,(a,b)}}=-\frac{b}{a}=\] the slope of \[\frac{x}{a}+\frac{y}{b}=2\] Hence, it touches the curve at (a, b) whatever may be the value of n.


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