The equation of the tangent to the parabola `y^(@)=4ax` at `(at^(2),2at)` is
`ty=x+at^(2)`
Here, a=2. So the equation of the tangent at `(2t^(2),4t)` to the parabola `y^(2)=8x` is
`ty=x+2t^(2)` (1)
The slope of (1) is 1/t and that of the given line is 3. Therefore,
`((1)/(t)-3)/(1+(1)/(t)xx3)=pmtan45^(@)=pm1`
i.e., `t=-(1)/(2)or2`
For t=-1/2, the tangent is `(-(1)/(2))y=x+2((1)/(4))`
i.e., `2x+y+1=0`
and the point of contact is (1/2,-2).
For t=2, the tangent is 2y=x+8 and the point of contact is (8,8).
