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JEE Main & Advanced Sample Paper JEE Main Sample Paper-31

  • question_answer
    Tangent to the ellipse \[\frac{{{x}^{2}}}{32}+\frac{{{y}^{2}}}{18}=1\] having slope \[\frac{-3}{4}\] meet the coordinate axes in A and B. The area of \[\Delta AOB\] (0 is origin) equals

    A) 12

    B) 8

    C) 24       

    D) 32

    Correct Answer: C

    Solution :

    Equation of tangent with slope \[=\frac{-3}{4}\]                                 \[y=\frac{-3}{4}x+C\] Now, \[C=\sqrt{32\times {{\left( \frac{-3}{4} \right)}^{2}}+18}=\sqrt{18+18}=6\] (Using condition of tangency) \[\therefore \]\[y=\frac{-3}{4}x+6\Rightarrow 3x+4y=24\] It meets the coordinate axes in A and B. So, A(8, 0) and B(0, 6). Hence, required area of \[\Delta AOB=\frac{1}{2}(8)(6)=24\]


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