Question

Find the equation of tangent to the curve x²+3y-3 = 0 which is parallel to the line y=4x-5 ​

Answer 1

Solution :- The given curve x²+3y-3 = 0 _____(1)

-(x²-3 )= 3y

dy/dx= 1/3(-2x) _____(3)

Slope of line , y = 4x-5

dy/dx = 4___(3)

Equate dy/dx of (1) and (2)

4 = 1/3(-2x)

x=-6

From (3) y = -1/3(x²-3)= -1/3(36-3)= -33/3= -11

tangent at (-6,-11) with slope=4 is

y + 11 = 4(X+6)

y = 4x+13 Answer .

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Answer 2

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Solution :- The given curve x²+3y-3 = 0 _____(1)

-(x²-3 )= 3y

dy/dx= 1/3(-2x)

Slope of line , y = 4x-5

By hypothesis , 4 = 1/3(-x²)

x=-6

From (1) y = 1/3(3-36) = -11

tangent at (-6,-11) with slope=4 is

y + 11 = 4(X+6)

y = 4x+13 Answer .