Locus and the Parabola Question (1 Viewer)

[Aron]

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

 

[HeroicPandas]

YOUR AIM IS TO EVALUATE

Remember if then

 

[Aron]

YOUR AIM IS TO EVALUATE

Remember if then

OK but that didn't really make things any clearer. Do i differentiate in respect to y?

 

[rumbleroar]

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]

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


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[Aron]

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.

 

[Drongoski]

 

[Aron]

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

 

[braintic]

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)

 

[Aron]

ok so i get (y+2)^2=16(x-3). now what?