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# c4 differentiation HELPPP watch

1. i really have no clue how to do part b) of this question, can someone help me please?

"the equation of the curve C is 6x^2y - 7 = 5x - 4y^2 - x^2
a) the line T has the equation y = c and passes through a point on C where x = 2. Find c, given that c > 0.

b) T also crosses C at the point Q
(i) find the coordinates of Q
(ii) find the gradient of C at Q."

i found the answer of 0.5 for part a, but i dont know what to do for part b), is x still 2 or has it changed???

2. (Original post by Nxsrinn)
i really have no clue how to do part b) of this question, can someone help me please?

"the equation of the curve C is 6x^2y - 7 = 5x - 4y^2 - x^2
a) the line T has the equation y=x and passes through a point on C where x = 2. Find c, given that c > 0.

b) T also crosses C at the point Q/
(i) find the coordinates of Q
(ii) find the gradient of C at Q."

i found the answer of 0.5 for part a, but i dont know what to do for part b), is x still 2 or has it changed???

Are you sure you've written the question correctly, in full?
3. (Original post by NotNotBatman)
Are you sure you've written the question correctly, in full?
sorryy i made a mistake, ive corrected it!
4. (Original post by Nxsrinn)
i really have no clue how to do part b) of this question, can someone help me please?

"the equation of the curve C is 6x^2y - 7 = 5x - 4y^2 - x^2
a) the line T has the equation y = c and passes through a point on C where x = 2. Find c, given that c > 0.

b) T also crosses C at the point Q
(i) find the coordinates of Q
(ii) find the gradient of C at Q."

i found the answer of 0.5 for part a, but i dont know what to do for part b), is x still 2 or has it changed???

Use the other (negative solution) of c. This is the y coordinate of q, back substitute to find the x coordinate of q.
5. (Original post by NotNotBatman)
Use the other (negative solution) of c. This is the y coordinate of q, back substitute to find the x coordinate of q.
okay! thank you for the help!!
6. (Original post by Nxsrinn)
okay! thank you for the help!!
No problem, are you okay with part (ii)?
7. (Original post by NotNotBatman)
No problem, are you okay with part (ii)?
yeah i think so, is it just implicit differentiation and then i sub in x = 2 and y = -13/2 ? for the gradient at that point ?
8. (Original post by Nxsrinn)
yeah i think so, is it just implicit differentiation and then i sub in x = 2 and y = -13/2 ? for the gradient at that point ?
Yes, that's it.

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