question_answer

The tangents to the curve \[y={{\left( x-2 \right)}^{2}}-1\]at its points of intersection with the line \[xy=3,\] intersect at the point : [JEE Main 12-4-2019 Afternoon]

A) \[\left( -\frac{5}{2},-1 \right)\]            

B) \[\left( -\frac{5}{2},1 \right)\]

C) \[\left( \frac{5}{2},-1 \right)\] 

D) \[\left( \frac{5}{2},1 \right)\]

Correct Answer: C

Solution :

Put \[x2=X\And y+1=Y\] \[\therefore \]given curve becomes \[Y={{X}^{2}}\]and \[Y=X\] tangent at origin is X-axis and tangent at \[A\left( 1,1 \right)\] is \[Y+1=2X\] \[\therefore \]there intersection is\[\left( \frac{1}{2},0 \right)\] \[\therefore \]\[x-2=\frac{1}{2}\And y+1=0\]therefore\[x=\frac{5}{2},y=-1\]