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Differential Calculus> tangents and normals find the equation of...

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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

taniska , 10 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 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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