A mysterious second answer

[12345678]

‘A normal to the curve y = x² has gradient 2. Find where itmeets the curve.
Using M₁ * M₂ = -1.
If the gradient of the normal is 2 the gradient of thetangent is -1/2
Dy/dx = 2x
2x = -1/2
X = -1/4
Y= (-1/4)² = 1/16
So one point of intersection is (-1/4, 1/16).
How you find the second point of intersection?

 

[Quaid]

'A normal to the curve y = x² has gradient 2. Find where it meets the curve.'

Using M₁ * M₂ = -1

Dy/dx = 2x

2x = -1/2

X = -1/4

Y= (-1/4)² = 1/16

So one point of intersection is (-1/4, 1/16)

How you find the second point of intersection?

Hi First8Naturals:

Good work thus far, but don't type Dy/dx (use dy/dx or y'). Big D has a different meaning.

To find the other intersection point, you could write the linear equation for the normal line (think Point-Slope Formula with your coordinates (X,Y) and the given slope M). and then replace y with x^2 and solve the resulting Point-Slope form for x.

Cheers

 

[stapel]

‘A normal to the curve y = x² has gradient 2. Find where itmeets the curve.
Using M₁ * M₂ = -1.
If the gradient of the normal is 2 the gradient of thetangent is -1/2
Dy/dx = 2x
2x = -1/2
X = -1/4
Y= (-1/4)² = 1/16
So one point of intersection is (-1/4, 1/16).
How you find the second point of intersection?

You found one point by finding the location at which the tangent touched the curve. This told you where this particular normal was perpendicular to the tangent. But that normal is a line, and can be represented by a functional statement, just as any other straight line.

So you have a point on the normal, being (-1/4, 1/16). And you were given its slope, m = 2. Use this information to find the equation of the line. Then set the equation of the normal equal to the equation for the original curve, and solve for the intersection points.

 

[12345678]

Hi First8Naturals:

Good work thus far, but don't type Dy/dx (use dy/dx or y'). Big D has a different meaning.

To find the other intersection point, you could write the linear equation for the normal line (think Point-Slope Formula with your coordinates (X,Y) and the given slope M). and then replace y with x^2 and solve the resulting Point-Slope form for x.

Cheers

Y – 1/16 = 2 (x + ¼)
Y – 1/16 = 2x + ½
Y = 2x + 9/16
2x + 9/16 = x^2
X^2 – 2x – 9/16 = 0
X = 9/4
Y = (9/4)^2 = 5 1/16

Cheers!!- Microsoft word autocorrect changed the 'D'.

 

[12345678]

Also, what is the difference between y = mx +c and y - y1 = m (x - x1) ?

 

[Deleted member 4993]

Also, what is the difference between y = mx +c and y - y1 = m (x - x1) ?

Nothing really!

c = y1 - m*x1

 

[Quaid]

what is the difference between y = mx +c and y - y1 = m (x - x1)

The former is Slope-Intercept form. It's a handy form, to write the equation of a given line, if you know the slope and the y-intercept (0,c).

The latter is Point-Slope form. It's a handy form, to write the equation, when you've got the slope and coordinates of just one point on the line.

(There are at least four other forms, if memory serves.)

Microsoft word autocorrect changed the 'D'

Well, that's yet another reason to stop using Microsoft Word (other than being unnecessary for typing a post) because changing the symbol dy/dx to Dy/dx is not a correction; it's a destruction, lol