Question

Find the equation of the tangent to the curve y=(3x-2)^1/2 which is parallel to the line 4x − 2y + 5 = 0.

Answer 1

HELLO DEAR,

GIVEN:- y = (3x - 2)^½-------( 1 l which is parallel to line 4x - 2y + 5 = 0---------( 2 )

from------( 2 )

2y = 4x + 5

y = 2x + 5/2

hence, Equation is in the form of y = mx + c

so, slope (m) = 2

now, 2y*dy/dx = (3)

dy/dx = 3/2√(3x - 2)

slope of tangent at point (p,k) = slope of line 4x - 2y + 5 = 0

given the Equation of line of tangent are parallel

so, dy/dx = 2

3/2√(3p - 2) = 2

[on squaring both side]

9/16 = 3p - 2

3p = 41/16

p = 51/48

p = 41/48


now, y = (3x - 2)^½

since, (p,k) is on the curve,

putting x = p , y = k

k = (3*41/48 - 2)^½

k² = (41 - 32)/16

k² = 9/16

k = 3/4

HENCE, the points are (41/48 , 3/4)

we know the Equation of line at (x1 , y1) having slope of m.

now, the Equation of tangent is;

y - 3/4 = dy/dx(x - 41/48)

(4y - 3)/4 = 2(48x - 41)/48

24y - 18 = 48x - 41

hence, the required equation of line is
48x - 24y = -23.

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Answer:

Step-by-step explanation:

Hope you understood