11th Class Mathematics Conic Sections Question Bank Critical Thinking

  • question_answer If the chord joining the points \[(at_{1}^{2},\ 2a{{t}_{1}})\] and \[(at_{2}^{2},\ 2a{{t}_{2}})\] of the parabola \[{{y}^{2}}=4ax\] passes through the focus of the parabola, then [MP PET 1993]

    A)            \[{{t}_{1}}{{t}_{2}}=-1\]            

    B)            \[{{t}_{1}}{{t}_{2}}=1\]

    C)            \[{{t}_{1}}+{{t}_{2}}=-1\]          

    D)            \[{{t}_{1}}-{{t}_{2}}=1\]

    Correct Answer: A

    Solution :

               \[\frac{(y-2a{{t}_{2}})}{(2a{{t}_{2}}-2a{{t}_{1}})}=\frac{x-at_{2}^{2}}{(at_{2}^{2}-at_{1}^{2})}\]; As focus i.e., (a, 0) lies on it,            Þ \[\frac{-2a{{t}_{2}}}{2a({{t}_{2}}-{{t}_{1}})}=\frac{a(1-t_{2}^{2})}{a({{t}_{2}}-{{t}_{1}})({{t}_{2}}+{{t}_{1}})}\] Þ \[-{{t}_{2}}=\frac{(1-t_{2}^{2})}{({{t}_{2}}+{{t}_{1}})}\]            Þ \[-t_{2}^{2}-{{t}_{1}}{{t}_{2}}=1-t_{2}^{2}\]  Þ \[{{t}_{1}}{{t}_{2}}=-1\].


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