derivative questions... (1 Viewer)

[littleboy]

1. a) Show that the tangent to P: y = ax^2 +bx + c with gradient m has y-intercept c - (m-b)^2/4a

b) Hence find the equations of any quadratics that pass through the origin and are tangent to both y= -2x - 4 and to y = 8x -49

c) Find also any quadratics that are tangent to y=-5x-10, to y= -3x-7 and to y=x-7.

2. Suppose y = ax^3 + bx^2 + cx + d is a cubic (so that a do not = 0 ). Show that every point in the plane lies on at least one tangent to this cubic.

3. Find the equation of the tangent to the parabola y= (x-3)^2 at the point T where x=a, find the coordinates of the x-intercept A and y-intercept B of the tangent, and find the midpoint M of AB. For what value of a does M coincide with T?

thanks

 

[KFunk]

1. a) Show that the tangent to P: y = ax^2 +bx + c with gradient m has y-intercept c - (m-b)^2/4a

y' = 2ax +b
m=2ax +b
x = (m-b)/2a when the gradient of the tangent is m

y= a([m-b]/2a)² +b[m-b]/2a +c
= (m² -2mb +b²)/4a + (2mb -2b²)/4a +c
=(m² - b²)/4a +c

hence the tangent with grandient m has the equation:

y - [ (m² -b²)/4a +c ] = m[ x + (b-m)/2a]

let x=0 to find y intercept

y = m(b-m)/2a + (m² -b²)/4a +c
= (2mb -2m²)/4a + (m² - b²)/4a +c
=(-m² +2mb -b²)/4a +c
=-(m-b)²/4a +c
hence y= c - (m-b)²/4a

 

[KFunk]

b) Hence find the equations of any quadratics that pass through the origin and are tangent to both y= -2x - 4 and to y = 8x -49

I've never tried one like this but I'll give it a shot.

If the quadratic passes through the origin then y = ax² +bx + c, where c=0
y= -2x-4, gradient = -2 and y intercept = -4 (1)
y=8x -49, gradient = 8 and y intercept = -49 (2)

combining (1) with the expression obtained in 1.a) [y= c - (m-b)²/4a] but now where c=0
-4 = -(-2-b)²/4a ----> (b+2)²/4 = 4a (3)

combining (2) with the expression obtained in a)
-49 = -(8-b)²/4a ----> (8-b)²/49 = 4a (4)

Using (3) and (4)
(b+2)²/4 = (8-b)²/49
(b+2)/2 = (8-b)/7
(here I'm slightly unsure about where to take ± so I'll just take one. If someone could correct me on this and show me the convention that'd be cool.)

±7(b+2) = 16 - 2b
either b = 2/9 or b=-6

using (b+2)² = 4a
16 = 16a ---> a =1 when b=-6 -----> quadratic y = x² -6x

using b=2/9 seems to give something that isn't an answer. ^ There's one of them in any case .

 

[Trev]

3. Find the equation of the tangent to the parabola y= (x-3)^2 at the point T where x=a, find the coordinates of the x-intercept A and y-intercept B of the tangent, and find the midpoint M of AB. For what value of a does M coincide with T?

y= (x-3)² T(a, (a-3)²)
y' = 2x-6 gradient at T = (2a-6)
y-(a-3)²= (2a-6)(x-a)
expand gives:
y=2ax-6x-a²+9 tangent at T.

X int A(-a²+9 , 0)
Y intercept B(0 , (a+3)/2)

∴Midpoint = [ (9-a²)/2, (a+3)/4 ]
Substitute points of midpoint into equation of tangent, and after algebra bashing...
[4a³-8a²-35a+75=0]
So M coincides with T when a=-3 or 2.5 (couldn't be bothered factorising so just used graphmatica)