## DLS 2 years ago Find the equation to the tangent to the curve x^2+3y-3=0 which is parallel to the line y=4x-5

1. DLS

$\LARGE \frac{dy}{dx}=2x+3=4$ $\LARGE x=-\frac{1}{2},y=\frac{11}{12}$

2. DLS

$\LARGE (y-\frac{11}{12})=4(x+\frac{1}{2})$

3. DLS

my attempt^

4. amistre64

why x=-1/2 ?

5. DLS

oops 1/2

6. amistre64

other than that, your process is good

7. DLS

12y-48x+13=0 is what im getting :O

8. amistre64

y = -5/4 y+5/4 = 4(x-1/2) this is a line equation ... so if form does not matter this would suffice

9. amistre64

otherwise, algrbrate it to your hearts content

10. DLS

isnt y 11/12?

11. amistre64

.5^2 + 3(.5) - 3 .25 + 1.5 - 3 1.75 - 3 = -1.25 = -5/4

12. amistre64

since x = 1/2 and NOT -1/2 you have to reevaluate the y you found

13. DLS

arent we substituting x=1/2 in the equation of curve?

14. amistre64

yes, which is what i just did

15. DLS

okay! thanks! got it

16. amistre64

$\frac14+\frac32-\frac31$ $\frac14+\frac64-\frac{12}4$ $\frac{7-12}4$

17. DLS

@amistre64 sorry,it was 3Y and not 3X that is how I got 11/12...

18. DLS

it is x squared so -1/2 and 1/2 both are same that is why i didnt change the y component

19. amistre64

ohh then your derivative might need some looking at: x^2+3y-3=0 2x +3y' = 0 y' = -2x/3 = 4 x = 12/-2 = -6

20. amistre64

(6)^2+3y-3=0 33+3y=0, when y = -11

21. amistre64

y+11 = 4(x+6) would be the tangent line

22. DLS

(-6,-11)

23. DLS

CORRECT NOW! :D

24. amistre64

yay! ;)