JEE Main 29 June 2022 Shift 2 Solved Paper

tan‌

π

4

=|‌

m−3

1+3m

‌

‌

|
⇒‌

m−3

1+3m

=±1
⇒‌

m−3

1+3m

=+1‌ and ‌‌

m−3

1+3m

=−1
⇒m−3=1+3m‌ and ‌m−3=−1−3m
⇒2m=−4‌ and ‌4m=2
m=−2‌ and ‌m=‌

1

2

We know, Equation of tangent to the parabola y2=4m is y=mx+‌

a

m

and point of contact is (‌

a

m2

,‌

2a

m

)
∴ Equation of tangent
y=−2x−‌

a

2

and y=‌

x

2

+2a
∴ Point of contact A and B are
A(‌

a

(−2)2

,‌

2a

−2

)=A(‌

a

4

,−a)
B(‌

a

(‌ 1 2 )2

,‌

2a

(‌ 1 2 )

)=B(4a,4a)
As points A, B and S are colinear so area of triangle formed by those 3 points are zero.

‌ Area of ‌△ABS=‌

1

2

|

‌ a 4	−a	1
4a	4a	1
a	0	1

|
=‌

a

4

(4a−0)+a(4a−a)+1(0−4a2)
=a2+3a2−4a2=0
∴ Area of triangle is independent of value of a.
So, for all value of a>0 (already given a must be greater than 0 ) point A,B and S will be collinear.