Question

can sumone please answer this! eq of curve y^2 +2x= 13 . eq of line 2y+x=k . Find the value of k for which the line is a tangent to the curve

Answer 1

first find the equation of all the tangent lines to the curve to compare with

Answer 2

y' is the slope of the tangent line at a given point and every tangent line at a given point (a,b) will take the form:

y = y'(a) (x-a)+b

either manipulate that into the line form given, or adjust the line form into a y= format to compare with

Answer 3

i dont understand . how to get it when there is x and y in the curve :(

Answer 4

can please explain one by one . im quite blur here :( please .

Answer 5

D[y^2 +2x= 13]

D[y^2] + D[2x] = D[13]

D[y^2] + 2D[x] = D[13]

what are the derivatives?

Answer 6

or if a derivative "with respect to x" is more to your likeings, then we can prolly write it up like this:\[\frac d{dx}(y^2) + 2\frac d{dx}(x) = \frac d{dx}(13)\]

Answer 7

okay thenn?

Answer 8

still blur but im trying to catch up . really weak in this

Answer 9

\[\frac {dy}{dx}2y + 2\frac {dx}{dx}1 = 0\]
since dx/dx=1, and dy/dx = y'\[2yy' + 2 = 0\]divide off the 2 to get\[yy'+1=0 \]and solve for y'

Answer 10

and what is your k?

Answer 11

how should i know, i havent even gotten to the k yet :/ ... what is y'?

Answer 12

no k is act 8.5

Answer 13

the slope of the family of lines: 2y+x=k is -1/2

when y' = -1/2 we have possible points to play with

yy' + 1 = 0

y' = -1/y; which means when y=2 we can define a point to use in the line equation

Answer 14

y^2 +2x= 13

x = (13-y^2)/2

x = 9/2; so a point on the line is (9/2, 2)

2(2) + 9/2 = k

Answer 15

so yes, k = 17/2, which is 8.5

Answer 16

oh yeah thanks a lot !! but i dont understand how u come with " yy' + 1 = 0"