Find equations to both lines through the point (0,-4) that are tangent to the parabola y=x^2-x.?

Question

"Well, first of all there is only one line that is tangent to the parabola at any particular point."

The point (0,-4) is not on the curve y=x^2-x, therefore there are two lines tangent to the curve from that point.

Answer 1

we don't know the points where the two lines going through the point (0,-4) cut the parabola y= x^2-x

let us call it the points (x,x^2-x)

if we differentiate y, we will find the gradient of the tangent at the point(x,x^2-x)

>>>>y'=2x-1

if the point (0,-4) = (x1,y1)

so, 2x-1 =(y-y1)/(x-x1)

=((x^2-x)-(-4))/(x-0)

>>> 2x-1= (x^2-x+4)/x

2x^2-x = x^2-x+4>>>>>x^2 = 4

x=2 or x= -2

therefore, substituting into y= x^2-x we have the

points (2,2) and (-2,6)

hence, the two tangent lines pass through the
common point (0,-4) and (2,2) and(-2,6)

let y1= -4,x1=0,y2 =2 and x2=2

the tangent line passing through the point (0,-4) and (2,2) is

(y-y1)=(y2-y1)(x-x1)/(x2-x1)

(y-(-4))= (2 -(-4))(x-0)/(2-0)

>>>> y=3x-4

let y1= -4,x1=0,y2 =6 and x2= -2

the tangent line passing through the point (0,-4) and (-2,6)
is

(y-(-4)) = (6-(-4))(x-0)/(-2-0)= 10x/(-2)= -5x

>>>>>> y= -5x-4

note if we substitute into y' at the points where the tangent lines through (0,-4) touch the parabola,we have

at x= -2, y'=2x-1 = -5 (y= -5x-4) and
at x=2, y'=2x-1= 3 (y= 3x-4)

if i could produce a drawing,the whole problem
would be a lot easier to explain

Answer 2

Suppose Charlie is a line that is tangent to the parabola at the point (a,b) and passes through the point (0,-4).

Note: the point (a,b) is on both the parabola and the line Charlie, but (0,-4) is a point only on the line Charlie.

Using the two known points on Charlie, we can find the slope of Charlie using the definition of slope: m = (b+4)/a.

Using Calculus we can find the slope of Charlie using the derivative of y = x^2 - x and the fact that (a,b) is on the line:

y' = 2x -1

so the slope of Charlie is m = 2a-1. Equating the two expressions yields

2a - 1 = (b+4)/a

Multiplying by a yields

2a^2 - a = b+4.

Now since (a,b) is on the parabola, we know that

a^2 - a = b.

Using the last two equations, we get

2a^2 - a = a^2 - a +4

Solving for a gives us

a^2 = 4

that is,

a = 2 or a = -2.

Therefore, the point (a,b) is either (2,2) or (-2,6). The corresponding tangent line equations are:

y - 2 = 3(x-2)

and

y + 2 = -5(x-6).

Answer 3

Supposing point (0,-4) is on the curve. Find the derivative of the parabola, substitute in 0 to find y' at x = 0, and use y-y1=m(x-x1) since you know x1 = 0, y1 = -4 and m = y' = -1. And yes, there is only one tangent line at a point.

If you mean both lines as in the tangent and the normal, then the normal line's slope is simply perpendicular (negative reciprocal) to the slope of the tangent line, -(1/-1) = 1. Now, use y-y1=m(x-x1) again to find the equation, only this time substitute 1 for m.

Again, I overlooked the problem, so it's wrong. But it's still easy to figure out the two tangent lines. Shamu is correct behind the reasoning.

Answer 4

Shamu is correct. The equations of the two lines are:

y =3x-4 and y= -5x -4

The slope of the tangent line to y = x^2 -x at point x is given by the derivative: dy/dx = 2x-1.

let (x , x^2-x) be the co-ordinates of the points on the parabola belonging to the tangent lines that pass through (0,-4). Then the slope of the lines is given by (y2-y1)/ (x2-x1), which in our case is ((x^2-x) - (-4)) / (x-0) = (x^2 -x+4)/x. This must be equal to the slope given by the derivative, which is 2x-1. So we must solve the equation (x^2-x+4)/x = 2x-1.

That is x^2 - x+4 = 2x^2 -x
x^2 -4 =0
implies x = 2 or x = -2

if x =2 then the slope = 2x-1 = 3
if x = -2 then the slope = 2x-1 = -5

Now we can determine the the two desired eqautions:

equ. 1

y = 3x +b Plugging in (0,-4) we have:

-4 = 3(0) + b, so b = -4 and equ 1 is:

y = 3x-4

Equ. 2

y = -5x+b Plugging in (0,-4) again gives b = -4.

So equ 2 is y = -5x -4

Answer 5

Well, first of all there is only one line that is tangent to the parabola at any particular point. So to find the equation of a straight line, we need two pieces of information:
-the gradient
-a point on the line

We have been given a point on the line, and to find the gradient, we use the equation of the parabola, the derivative of a function essentially does that, it gives us the slope of the line at that particular point. so with:

y=x^2-x, we differentiate to get:

y'=2x-1, we then substitute the values of the points given into that, namely we substitute (0,-4), into y', and we get y'=gradient=m=-1.

Then, to find the equation of the straigh line, we use the point-gradient formula; y-y1=m(x-x1), and we substitute the value of the gradient in m, and the point we were given into the equation; (0,-4)=(x1,y1).

There we go.

Answer 6

y = x^2 - x and so gradient = dy/dx = 2x - 1

Let the point on the curve where the tangent touch be at x=r.
The point that the tangent pass through is (0, -4).
The coordinates are: (r, r^2-r) and the gradient is: 2r-1.

So, [r^2-r-(-4)]/[r-0] = 2r-1
r^2-r+4 = 2r^2-r
4 = r^2
r = 2 or -2

So the coordinates are:
(2, 2^2-2) and (-2, (-2)^2-(-2))
= (2, 2) and (-2, 6)

And the gradients are: 2(2)-1 = 3 and 2(-2)-1 = -5

Substituting (2,2) and m=3,
y=mx+c
2=3(2)+c
2-6=c
c=-4
So y = 3x-4

Substituing (-2,6) and m=-5,
y=mx+c
6=(-5)(-2)+c
6=10+c
6-10=c
c=-4
So y = -5x-4

Answer 7

the point on y can be expressed as (p, p^2-p),

the slope of y at any point is 2p-1,

slope between (0,-4) and (p,p^2-p) => p^2-p+4/p = 2p-1

p = 2, -2

when p = 2, slope = 3, the point on y is (2,2), equation y = 3x-4
wheh p =-2, slope = -5, the point on y is (-2,6) equation y = -5x -4