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
