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Find the equation of the tangent line to the curve `"y" = sqrt(5"x" -3) -5`, which is parallel to the line `4"x" - 2"y" + 5 = 0`.

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#### Solution

Here `"y" = sqrt(5"x" -3)-5`.

`dy/dx = 5/(2sqrt(5x - 3))`

Slope of line 4x - 2y + 5 = 0 is `(- 4)/(- 2) = 2`

∴ `5/(2sqrt(5x - 3)) = 2 x 73/80`

Putting x = `73/80 "in equation (i) we get y" = -15/4`

**Hence the equation of tangent: **

`"y"+(15)/(4) = 2 ("x" -(73)/(80))`

⇒ `80"x" - 40"y" = 223`.

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