The normal at the point (bt12, 2bti) on a parabola meets the parabola again at the points (bt22, 2bt2), then
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Answer:
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Step-by-step explanation:
Correct option is A)
y
2
=4bx
⇒
dx
dy
=
y
2b
Therefore, at the point (bt
1
2
,2bt
1
),
dx
dy
=
t
1
1
So, the slope of normal at (bt
1
2
,2bt
1
)=
(
dx
dy
)
−1
=−t
1
Therefore, the equation of normal at (bt
1
2
,2bt
1
) is (y−2bt
1
)=−t
1
(x−bt
1
2
)
The point (bt
2
2
,2bt
2
) also lies on the normal
⟹(2bt
2
−2bt
1
)=−t
1
(bt
2
2
−bt
1
2
)
⟹2b(t
2
−t
1
)=−t
1
b(t
2
−t
1
)(t
2
+t
1
)
⟹2b(t
2
−t
1
)+t
1
b(t
2
−t
1
)(t
2
+t
1
)=0
⟹b(t
2
−t
1
)(2+t
1
(t
2
+t
1
))=0
⟹2+t
1
(t
2
+t
1
)=0 ....[t
2
=t
1
] and [b
=0]
⟹t
2
=−t
1
−
t
1
2
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