2. Solved Papers
  3. BCECE Engineering

BCECE Engineering BCECE Engineering Solved Paper-2006

  • question_answer
    If \[{{M}_{1}}\]and \[{{M}_{2}}\]are the feet of the perpendiculars from the foci \[{{S}_{1}}\]and \[{{S}_{2}}\]of the ellipse\[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}=1\] on the tangent at a point P on the ellipse, then \[({{S}_{1}}{{M}_{1}})({{S}_{2}}{{M}_{2}})\]is equal to:

    A)  16                         

    B)         9                            

    C)  4                            

    D)         3

    Correct Answer: B

    Solution :

    Key Idea: The product of perpendiculars from the foci on any tangent to the ellipse\[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] is equal to \[{{b}^{2}}.\] We know that the product of perpendiculars from two foci \[{{S}_{1}}\]and \[{{S}_{2}}\]of an ellipse \[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}=1.\]on the tangent at any point P on  the ellipse is equal to the square of the semi-minor axis.                 \[({{S}_{1}}{{M}_{1}})({{S}_{2}}{{M}_{2}})=9.\]


You need to login to perform this action.
You will be redirected in 3 sec spinner