JEE Main & Advanced Sample Paper JEE Main Sample Paper-6

  • question_answer
    If the chords of contact of tangents from two points \[({{x}_{1}},{{y}_{1}})\] and \[({{x}_{2}},{{y}_{2}})\] to the hyperbola \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\]are at right angles, then \[\frac{{{x}_{1}}{{x}_{2}}}{{{y}_{1}}{{y}_{2}}}\] is equal to

    A)  \[-\frac{{{a}^{2}}}{{{b}^{2}}}\]                  

    B)  \[-\frac{{{b}^{2}}}{{{a}^{2}}}\]

    C)  \[-\frac{{{b}^{4}}}{{{a}^{4}}}\]                  

    D)  \[-\frac{{{a}^{4}}}{{{b}^{4}}}\]

    Correct Answer: D

    Solution :

    The equation of the chords of contact of tangents from \[({{x}_{1}},{{y}_{1}})\] and \[({{x}_{2}},{{y}_{2}})\]to the given hyperbola are\[\frac{x{{x}_{1}}}{{{a}^{2}}}-\frac{y{{y}_{1}}}{{{b}^{2}}}=1\]           ...(i) and\[\frac{x{{x}_{2}}}{{{a}^{2}}}-\frac{y{{y}_{2}}}{{{b}^{2}}}=1\]                                               ?(ii) Equation of lines (i) and (ii) are at right angled. \[\therefore \]\[\frac{{{b}^{2}}}{{{a}^{2}}}\frac{{{x}_{1}}}{{{y}_{1}}}\times \frac{{{b}^{2}}}{{{a}^{2}}}\frac{{{x}_{2}}}{{{y}_{2}}}=-1\]\[\Rightarrow \]\[\frac{{{x}_{1}}{{x}_{2}}}{{{y}_{1}}{{y}_{2}}}=-\frac{{{a}^{4}}}{{{b}^{4}}}\]


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