+0

# Pls explain. Thx u very much!

0
42
2

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.

Sep 26, 2023

#1
+129396
+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)

Sep 26, 2023
#2
+36
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 * 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.

Sep 27, 2023