Locus and the Parabola Question (1 Viewer)

Aron

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Challenge Question 12

Find the equation of the tangent to the parabola y^2+4y-16x+52=0 at the point where x=4 in the first quadrant. answer is 2x-y-6

Thanks in Advance! =D
 
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rumbleroar

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Make y the subject by square rooting the other side OR use implicit differentiation. Then sub x=4 into the equation and solve for the gradient and tangent.


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rumbleroar

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Wait sorry I don't think you can make y the subject, so you must implicitly differentiate it.


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Aron

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Wait sorry I don't think you can make y the subject, so you must implicitly differentiate it.
So what does differentiating y^2+4y-16x+52=0 give? I'm having trouble making y the subject.
 

Aron

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The answer is 2x-y-6. Plus the information talks about x=4 so i don't know where to fit that in.
 

braintic

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Wait sorry I don't think you can make y the subject, so you must implicitly differentiate it.


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Yes you can! Move the 16x term to the other side and complete the square. You'll get a plus/minus, but you can discard the minus given that the point is in the first quadrant.

Implicit differentiation is not in the 2 unit course.

(Having said that, implicit differentiation is easier)
 

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