if t1 t2 t3 are distinct the points (t1,2at1,at1^3 )(t2,2at2,at2^3)and (t3,2at3,at3^3) are collinear then show that t1^3+t2^3+t^3=3t1t2t3
Answers
Answer:
dont know soory
Explanation:
Answer:
Correct option is
A
t
1
t
2
t
3
=−1
Let points are P(t
1
.2at
1
+at
1
3
) ,Q(t
1
,2at
2
+at
1
3
) and R(t
3
,2at
3
+at
3
3
)
P,Q and R are collinear
⇒area ( ΔPQR) =0
⇒
∣
∣
∣
∣
∣
∣
∣
∣
1
1
1
t
1
t
2
t
2
2at
1
+at
1
3
2at
2
+at
2
3
2at
3
+at
3
3
∣
∣
∣
∣
∣
∣
∣
∣
=0
⇒
∣
∣
∣
∣
∣
∣
∣
∣
1
1
1
t
1
t
2
t
3
2at
1
2at
2
2at
3
∣
∣
∣
∣
∣
∣
∣
∣
+
∣
∣
∣
∣
∣
∣
∣
∣
1
1
1
t
1
t
2
t
3
at
1
3
at
2
3
at
3
3
∣
∣
∣
∣
∣
∣
∣
∣
=0
⇒2a
∣
∣
∣
∣
∣
∣
∣
∣
1
1
1
t
1
t
2
t
3
t
1
t
2
t
3
∣
∣
∣
∣
∣
∣
∣
∣
+a
∣
∣
∣
∣
∣
∣
∣
∣
1
1
1
t
1
t
2
t
3
t
1
3
t
2
3
t
3
3
∣
∣
∣
∣
∣
∣
∣
∣
=0
Determinant is zero if two rows of a determinant are same
⇒0+a
∣
∣
∣
∣
∣
∣
∣
∣
1
1
1
t
1
t
2
t
3
t
1
3
t
2
3
t
3
3
∣
∣
∣
∣
∣
∣
∣
∣
=0
R
2
→R
2
−R
1
R
3
→R
3
−R
1
⇒
∣
∣
∣
∣
∣
∣
∣
∣
1
1
0
t
1
t
2
−t
1
t
2
−t
1
t
1
3
t
2
3
−t
1
3
t
3
3
−t
1
3
∣
∣
∣
∣
∣
∣
∣
∣
=0
⇒(t
2
−t
1
)(t
3
−t
1
)
∣
∣
∣
∣
∣
∣
∣
∣
1
0
0
t
1
1
1
t
1
3
t
1
2
+t
1
t
2
+t
2
2
t
1
2
+t
1
t
3
+t
3
2
∣
∣
∣
∣
∣
∣
∣
∣
=0
⇒(t
2
−t
1
)(t
2
−t
1
)[t
1
2
+t
!
t
2
+t
3
2
−t
1
2
−t
1
t
2
−t
2
2
]=0
⇒(t
2
−t
1
)(t
3
−t
1
)[t
2
3
+t
1
t
3
−t
1
t
2
−t
2
2
]=0
⇒(t
2
−t
1
)(t
3
−t
1
)(t
3
−t
2
)(t
1
+t
2
+t
3
)=0
as t
1
=t
2
=t
3
⇒t
2
−t
1
or t
3
−t
1
or t
3
−t
2
can't be zero
⇒t
1
+t
2
+t
3
=0
Explanation: