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प्रश्न
Find the equation of tangent to the curve `y = sqrt(3x -2)` which is parallel to the line 4x − 2y + 5 = 0. Also, write the equation of normal to the curve at the point of contact.
उत्तर
Slope of the given line is 2
Let (x1, y1) be the point where the tangent is drawn to the curve `y = sqrt(3x -2)`
Since, the point lies on the curve.
Hence, `y_1 = sqrt(3x_1-2)` ...(1)
Now, `y = sqrt(3x -2)`
⇒ `dy/dx = 3/(2sqrt(3x-2)`
Slope of tangent at `(x_1,y_1) = 3/(2sqrt(3x_1-2)`
Given that
Slope of tangent = slope of the given line
⇒ `3/(2sqrt(3x_1-2)` = 2
⇒ `3 = 4 sqrt(3x_1-2)`
⇒ `9 = 16 (3x_1 - 2)`
⇒ `9/16 = 3x_1 -2`
⇒ `3x_1 = 9/16 + 2 = (9+32)/16 = 41/16`
⇒ `x_1 = 41/48`
Now,
`y_1 = sqrt(123/48 -2) = sqrt(27/48) = sqrt(9/16) = 3/4` ...[From (1)]
∴ `(x_1,y_1) = (41/48,3/4)`
Equation of tangent is,
`y - y_1 = m (x -x_1)`
⇒ `y - 3/4 = 2 (x-41/48)`
⇒ `(4y-3)/4 = 2 ((48x -41)/48)`
⇒ 24y - 18 = 48x - 41
⇒ 48x - 24y - 23 = 0
Equation of normal at the point of contact will be
`y-y_1 = -1/m (x -x_1)`
⇒ `y - 3/4 = -1/2 (x -41/48)`
⇒ `(4y-3)/4 = -1/2 (x -41/48)`
⇒ `(4y-3)/2 = (41/48 -x)`
⇒ `(4y-3)/2 = (41-48x)/48`
⇒ `4y - 3 = (41 -48x)/24`
⇒ 96y - 72 = 41 - 48x
⇒ 48x + 96y = 113
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