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    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???

    please help , Thank you!!
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    (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???

    please help , Thank you!!
    Are you sure you've written the question correctly, in full?
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    (Original post by NotNotBatman)
    Are you sure you've written the question correctly, in full?
    sorryy i made a mistake, ive corrected it!
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    (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???

    please help , Thank you!!
    Use the other (negative solution) of c. This is the y coordinate of q, back substitute to find the x coordinate of q.
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    (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!!
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    (Original post by Nxsrinn)
    okay! thank you for the help!!
    No problem, are you okay with part (ii)?
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    (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 ?
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    (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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