question_answer

At what point on the curve \[y={{x}^{2}}\] does the tangent make an angle of \[45{}^\circ \] with the X-axis?

Answer:

Given equation of curve is Now,\[y={{x}^{2}}\] Slope of the tangent \[=\tan 45{}^\circ =1\] \[y'=2x\] \[\Rightarrow \]\[2x=1\Rightarrow x=\frac{1}{2}.\] On putting the value of x in Eq. (i), we get \[y=\frac{1}{4}\]Hence, the required point is \[\left( \frac{1}{2},\,\,\frac{1}{4} \right)\].