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Given that , we have to find tangent to curve which is parallel to the line 4x-2y+5=0.

Differentiating y with respect to x, we get  Now at (x1,y1) the equation will be: The slope of the tangent at (x1,y1 ) is: Now,

The slope of the tangent = slope of the line  Squaring both sides we get,

9 = 16(3x1 - 2)    Now,      Equation of tangent is:

y - y1 = m (x - x1)  ⇒ 6(4y - 3) = 48x - 41

⇒ 24y - 18 = 48x - 41

⇒ 48x - 24y - 23 = 0

Equation of normal is:     96y - 72 = 41 - 48x

⇒48x + 96y - 113 = 0

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