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KVPY Sample Paper KVPY Stream-SX Model Paper-25

  • question_answer
    The circumcentre of a triangle lies at the origin  and its centroid is the mid-point of the line segment joining the points \[\left( {{a}^{2}}+1, {{a}^{2}}+1 \right)\] and \[(2a,-2a),a\ne 0.\] Then for any a, the orthocentre of this triangle lies on the line:

    A) \[\operatorname{y}-2ax=0\]

    B) \[y-\left( {{a}^{2}}+1 \right)x=0\]

    C) \[~y+x=0\]

    D) \[{{\left( \operatorname{a}-1 \right)}^{2}}x-{{\left( \operatorname{a}+1 \right)}^{2}}y=0\]

    Correct Answer: D

    Solution :

    circumcentre \[=(0,0)\]
    Centroid \[=\left( \frac{{{\left( a+1 \right)}^{2}}}{2},\frac{{{\left( a-1 \right)}^{2}}}{2} \right)\]
    We know the circumcentre (O),
    Centroid (G) and orthocentre (H) of a triangle lie on the line joining the O and G.
    Also, \[\frac{HG}{GO}=\frac{2}{1}\]\[\Rightarrow \] Coordinate of orthocentre \[=\frac{3{{(a+1)}^{2}}}{2},\frac{3{{(a-1)}^{2}}}{2}\]
    Now, these coordinates satisfies eqn given in option [d]
    Hence, required eqn of line is\[{{\left( a-1 \right)}^{2}}x-{{\left( a+1 \right)}^{2}}y=0\]


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