=−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.