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The equation of tangent at (2, 3) on the curve y2 = ax3 + b is y = 4x – 5. Find the values of a and b. - Mathematics

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The equation of tangent at (2, 3) on the curve y2 = ax3 + b is y = 4x – 5. Find the values of a and b.

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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=2ax4a+3      .....(2)

But according to question,

Equation of tangent at the point (2,3) is y=4x5

Both the equation represents the same line, therefore comparing the coefficients of both the line, we have

2a=4a=2 and 34a=5a=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.

Concept: Tangents and Normals
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