UPSC NDA I 2025 Mathematics Paper

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)