Finding an equation of the normal to curves (2)

[Monkeyseat]

Ok, I would appreciate if you could check this question I have done. I believe I have done it correctly but the answers say otherwise. One part of the question was to find the value of one y co-ordinate (for point P below) and the length of PQ so I have included all the info and previous answers in the question in case it is needed.

Question:

The points P (-1, 11) and Q (4, 6) lie on the curve with the equation y = x^2 - 4x + 6. PQ is length 5(sqr. 2). Find an equation for the normal to the curve at the point Q.

My working

Curve: y = x^2 - 4x + 6, Point: (4, 6)

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dy/dx = 2x - 4

When x = 4, dy/dx = (2 * 4) - 4 = 4

Gradient of normal:

m1 * m2 = -1
4 * (-1/4) = -1

Therefore the gradient of the normal = -1/4

Equation of normal (x1 = 4, y1 = 6, m = -1/4):

y - y1 = m(x - x1)
y - 6 = -1/4(x - 4)
4(y - 6) = -(x - 4)
4y - 24 = -x + 4
4y + x - 28 = 0

For this Question, the book says the answer is 4y = x + 10, who is correct?

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I have done a lot of these questions and this is the only 1 I have done wrong apparently. If you could check it and tell me where I'm going wrong it would be appreciated (preferably using the methods I have learnt) as I have checked many times and don't know what error I have made. The book makes quite a few mistakes however, so I wasn't sure. Any typos let me know.

Thanks!

 

[stapel]

I get your answer, not the book's.

Eliz.

 

[Monkeyseat]

Once again stapel many thanks. First post meant to say PQ is length 5(sqr. 2) not necessarily the whole curve by the way (edited).