The graph of y =\( {1 \over 2}\)(3x2 - k) has a normal with the equation 6y + 2x = -1 at point A.
(i) Show that k = 4.
The tangent to the curve at point A cuts the y-axis at B.
(ii) Find the coordinates of B.
+129885 6y + 2x = -1
6y = -2x -1
y = (-1/3)x - 1/6
The slope of this line = -1/3
y= (1/2)(3x^2 - k)
y= (3/2)x^2 - k/2 the derivative of this is
y' = 3x
We are looking for the point on this function where the the slope = 3
So
3x = 3
x =1
When x = 1, y =(-1/3)x - 1/6 = (-1/3)(1) - 1/6 = -1/3 - 1/6 = -1/2
So....to find k....
-1/2 = (1/2)(3(1)^2 - k)
-1/2 = 3/2 - k/2
-4/2 = -k/2
4 = k
Second part
The equation of the tangent line at A = (1, -1/2) is
y = 3(x -1) -1/2
y = 3x - 3 -1/2
y= 3x - 7/2
This line will intercept the x axis when y = 0
0 = 3x -7/2
7/2 = 3x
x = 7/6
B= (7/6 , 0)
+36 The normal has an equation
6y + 2x = -1 (given).
Its slope is -1/3.
It means that the tangent line to the parabola has the slope m= 3. From the given equation of the parabola, the slope of the tangent line at abscissa x is (1/2) * 3 * 2x = 3x (after taking the derivative).
In order for the slope of a tangent line 3x be m= 3, x should be equal to 1.
Thus we found out that x-coordinate of point A is x= 1.
Substitute x= 1 into equation of the normal line 6y + 2x = -1 . You will get 6y + 2*1 = -1, or 6y = -1 - 2 = -3, which gives y = -3/6 = -1/2.
Thus we learned that tangent point A on the parabola has coordinates A = (x,y) = (1, -1/2).
Now we know that the parabola passes throw the point (1, -1/2).
From equation of the parabola, it means that
-1/2 = (1/2) * (3 * 12 - k).
It gives
-1 = 3 - k,
which implies
k = 3 + 1 = 4.
k = 4.
Part ( i ) is completed in full.
Part ( ii ) is simple arithmetic.
