Aye,
I usually get questions with tangents and normals but this one confuses me o.0
The normal to the curve y=x^2 +1 at the point where x= 2 ( first part, finding the normal) cuts the curve again at point P ( second part, find intersections, ie solve simultaneously) . Find the coordinate of p.
Working out would be nice thanks
so we find eqn of normal, normal is perpendicular to tangent
so y=x^2 +1
dy/dx= 2x
when x= 2, dy/dx= gradient of tangent= 4
so as m1 * m2 =-1 for perpendicular lines
so m of normal = -1/4
when x=2, y= (2)^2 +1= 5
(2, 5) m= -1/4
y-y1=m(x-x1)
y-5=-1/4 ( x-2)
y= -x/4 +11/2
now to find intersections of this normal with the curve is equivalent to solving simultaneously the eqns
y= -x/4 +11/2 ---eqn1
y= x^2 +1 --- eqn 2
they both equal y, so set them equal to each other
x^2 +1 = -x/4 +11/2
x^2 +x/4 -9/2=0
x= [ -b +- sqrt ( b^2 -4ac) ]/ [2a] { one of the solutions will be x=2 ( the one we already have), so take the other x value}
= [-1/4 +- sqrt(1/2 -4(1)(-9/2) ) ]/ 2
etc, then find the y coordinate by subing that x coordinate into either of the two eqns above
