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Question

If the curve y2=ax3−6x2+b passes through (0,1) and has its tangent parallel to y-axis at x=2, then

A
a=2,b=1
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B
a=238,b=1
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C
a=823,b=1
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D
a=238,b=1
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Solution

The correct option is A a=238,b=1
Given equation of curve
y2=ax36x2+b
Since, the curve passes through (0,1)
b=1
Since, the tangent is parallel to y-axis at x=2
y=8a23
So, point of tangency is (2,8a23)
Slope of tangent to curve
2ydydx=3ax212x
dydx=3ax212x2y
Slope of tangent to curve at (2,8a23) is 12a2416a46
Since, the tangent is parallel to y-axis
12a2416a46=10
16a46=0
a=238

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