Find the point on the curve `y^2=a x` the tangent at which makes an angle of 45^0 with the x-axis.

Answer

(a/4, a/2)

Explanation

The equation of any tangent to the parabola `Y^(2)=4ax` in terms of its slope m is
`y=mx+(a)/(m)`
The coordinates of the point of contact are `(a//m^(2),2a//m)`.
Therefore, the equation of tangent to `y^(2)=ax` is
`y=mx+(a)/(4m)`
and the coordinates of the point of contact are `(a//4m^(2),a//2m)`.
It is given that `m=tan45^(@)=1`.
Therefore, the coordinates of the point of contact are `(a//4,a//2)`.