CGPET 2021 Solved Paper

Given,
Parabola y2=4x,4a=4⇒a=1
Let point on the curve P(at2,2at),i.e.P(t2,2t)
Equation of tangent a point P(1,2) is
(y−2)=m(x−1)
m=

dy

dx

|(1,2)
⇒ y2=4x⇒2y

dy

dx

=4
⇒

dy

dx

=

2

y

⇒

dy

dx

|(1,2)=

2

2

=1
∴ Equation of tangent is
(y−2)=1(x−1)
x−y+1=0
Let the mirror image of point P(t2,2t)be(h,k)

h−t2

1

=

k−2t

−1

=

−2(t2−2t+1)

12+12

⇒

h−t2

1

=

k−2t

−1

=

−2(t2−2t+1)

2

⇒

h−t2 1

I

=

k−2t −1

II

=

−t2+2t−1 1

III

Comparing I and III
h=2t−1⇒t=

h+1

2

|
Comparing II and III
k=t2+1⇒k−1=t2
∴ Eliminating t
k−1=(

h+1

2

)2
⇒ (h+1)2=4(k−1)
Put h=x,k=y, we get (x+1)2=4(y−1)