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tangents and normals find the equation of the tangent to the curve y= sqrt(3x – 2) which is parallel to the line 4x – 2y + 5 = 0

tangents and normals find the equation of the tangent to the curve y= sqrt(3x – 2) which is parallel to the line 4x – 2y + 5 = 0

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 years ago
Ans:

Hello Student,
Please find answer to your question,

y = \sqrt{3x-2}
Slope of tangent = 2 (Given)
y' = \frac{3}{2\sqrt{3x-2}}
\frac{3}{2\sqrt{3x-2}} = 2
3 = 4\sqrt{3x-2}
9 = 16(3x-2)
x = \frac{41}{48}
y = \sqrt{3\frac{41}{48}-2}
y = \sqrt{\frac{9}{16}} = \frac{3}{4}
You have the point and slope → line of equation.

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