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Question
The equation of tangent at (2, 3) on the curve y2 = ax3 + b is y = 4x – 5. Find the values of a and b.
Solution
The equation of the given curve is
y2=ax3+b .....(1)
Differentiate equation (1) with respect to x, we get
`2y dy/dx=3ax^2`
`dy/dx=(3ax^2)/(2y) `
`(dy/dx)_(2,3)=((3ax^2)/(2y))_(2,3)`
`=> (dy/dx)_(2,3)=2a`
So, equation of tangent at the point (2, 3) is
(y-3)=2a(x-2)
⇒y=2ax−4a+3 .....(2)
But according to question,
Equation of tangent at the point (2,3) is y=4x−5
Both the equation represents the same line, therefore comparing the coefficients of both the line, we have
2a=4⇒a=2 and 3−4a=−5⇒a=2 .....(3)
The point (2, 3) lies on the curve y2=ax3+b, so
(3)2=a(2)3+b
⇒9=8a+b
⇒9=8×2+b [From (3)]
⇒b=−7
Hence, the values of a and b are 2 and −7, respectively.
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