Gradient of norm question

[bluefrog]

I have attempted the following question.

Find the coordinates of the point where the tangent to the curve $y=x^2+1$ at the point $(2,5)$ meets the normal to the same curve at the point $(1,2)$

My attempt is as follows:
$$\frac{dy}{dx} (x^2+1)= 2x \\ \text{gradient: } 2(2) = 4 \\ \text{ Tangent: } y= 4x+c \Rightarrow 5 = 4(2) + c \Rightarrow y = 4x - 3 \\ \text {Norm: } y=-\frac{1}{4}x+c \Rightarrow 2=-\frac{1}{4}(1) +c \Rightarrow 2+1/4=c \Rightarrow y=-\frac{1}{4}x+\frac{9}{4} \\ \\ \text{Intersect: } \\ 4x-3=-\frac{1}{4}x+\frac{9}{4} \\ 4(4x-3)=-x+9 \\ 16x-12=-x+9 \\ 17x-21=0 \\ x=\frac{21}{17} \Rightarrow y = 4(\frac{21}{17}) -3=\frac{33}{17} \\ \Rightarrow (x,y)=(\frac{21}{17}, \frac{33}{17})$$
But the answer in the book is shown as, answer is at 3



No working out is shown, only the final answer is shown.
Can anybody suggest where I have gone wrong ?

 

[bluefrog]

I think I understand the question a bit better.

I got confused with what tangent the norm should be of. So clearly there are two tangents, the other being $y=2x$,
therefore the norm is $y=-\frac{1}{2}x+c$,
hence $2=-\frac{1}{2}x+c \Rightarrow c=2\frac{1}{2} \Rightarrow y=-\frac{1}{2}x+2\frac{1}{2}$

The intersect is therefore $4x-3=-\frac{1}{2}x+2\frac{1}{2}$, which when continued resolves to the same in the book, as shown above.