11th Class Mathematics Conic Sections Question Bank Critical Thinking

  • question_answer The centre of the circle passing through the point (0, 1) and touching the curve \[y={{x}^{2}}\]at (2, 4) is       [IIT 1983]

    A)            \[\left( \frac{-16}{5},\ \frac{27}{10} \right)\]                             

    B)            \[\left( \frac{-16}{7},\ \frac{5}{10} \right)\]

    C)            \[\left( \frac{-16}{5},\ \frac{53}{10} \right)\]                             

    D)            None of these

    Correct Answer: C

    Solution :

               Tangent to the parabola \[y={{x}^{2}}\]at \[(2,4)\]is                   \[\frac{1}{2}(y+4)=x.2\]or \[4x-y-4=0\]                   It is also a tangent to the circle so that the centre lies on the normal through (2, 4) whose equation is \[x+4y=\lambda ,\]where \[2+16=\lambda \]                   \[\therefore x+4y=18\]is the normal on which lies \[(h,k)\].                   \\[h+4k=18\]                                                 .....(i)                   Again distance of centre \[(h,k)\]from \[(2,4)\]and \[(0,1)\]on the circle are equal.                   \[\therefore {{(h-2)}^{2}}+{{(k-4)}^{2}}={{h}^{2}}+{{(k-1)}^{2}}\]                   \[\therefore 4h+6k=19\]                                                            .....(ii)                    Solving (i) and (ii), we get the centre\[=\left( \frac{-16}{5},\frac{53}{10} \right)\].

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