The graph of y =\( {1 \over 2}\)(3x^{2} - 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*.

Guest Sep 26, 2023

#1**+1 **

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)

CPhill Sep 26, 2023

#2**0 **

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 * 1^{2} - 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.

tastyabananas2ndDad Sep 27, 2023