Given the parabola y2=4x and a tangent to this parabola that is inclined at an angle of 45∘ with the positive x -axis, Hence, its slope is m=tan(45∘)=1. A standard parametric form for y2=4x is (x(t),y(t))=(t2,2t) since y2=4x⟹(2t)2=4t2 The slope of the tangent at (t2,2t) Differentiate y2=4x 2y
dy
dx
=4 ⇒
dy
dx
=
4
2y
=
2
y
. At the point (x,y)=(t2,2t) one has y=2t
dy
dx
|(t2,2t)=
2
2t
=
1
t
. We require this slope to equal 1 .
1
t
=1⇒t=1 Now point of contact Substitute t=1 in (x(t),y(t))=(t2,2t). x(1)=12=1, y(1)=2⋅1=2. Thus the point of contact of the tangent of slope 1 is (1,2)