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avatar+639 

The point A lies on the curve with the equation = 𝑥^1/2 . The tangent to this curve at A is parallel to the line 3𝑦 −2𝑥 = 1 . Find the equation of this tangent at A.

 Dec 2, 2018
 #1
avatar+94545 
+1

The line will have the slope   = 2/3

The derivative  of x^(1/2)   =   1 / ( 2√x)

 

So....to find the x coordiniate of A

 

2/3 = 1 / ( 2√x)

( 2√x)  = 3/2         square both sides

4x = 9/4

x = 9/16

 

So....the y coordinate of A is   (9/16)^1/2  =  3/4

 

And the equation of the tangent line is 

 

y = (2/3) (x - 9/16) + 3/4

y = (2/3)x  - 9/24 + 3/4

y = (2/3)x + 9/24

24y = 16x + 9

16x - 24y = -9

 

 

cool cool  cool

 Dec 2, 2018
 #2
avatar+639 
+1

can u explain how u got the slope please

YEEEEEET  Dec 2, 2018
 #3
avatar+94545 
+1

Oh yeah.....

 

In the form 

 

Ax + By  = C

 

The slope is given by   -A /B

 

So....the line   -2x + 3y = 1  has the slope    =  - (-2) / 3  =  2/3

 

Hope that helps

 

 

cool cool cool

CPhill  Dec 2, 2018
 #4
avatar+639 
+1

ye i always struggle to get the slope for those kinds thank you!

YEEEEEET  Dec 2, 2018

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