Find the equation of the normal to the curve y=2x^2+3x+4 which is perpendicular to the line y=7x-5
To find the equation of the normal to the curve that is perpendicular to the line, we need to find the slope of the line and then determine its negative reciprocal.
First, let's find the slope of the line y=7x-5 by comparing it to the standard slope-intercept form, y=mx+b. In this case, m represents the slope. So, the slope of the line is 7.
To find the slope of the curve y=2x^2+3x+4, we differentiate the equation with respect to x.
Differentiating y=2x^2+3x+4:
dy/dx = 4x + 3
The slope of the curve at any given point is given by dy/dx. So the slope of the curve y=2x^2+3x+4 is 4x + 3.
To find the equation of the normal, we need to find the negative reciprocal of the slope of the line.
For a line that is perpendicular to a given line, the negative reciprocal of the slope of the line is taken as the slope of the normal line.
So, the slope of the normal line is -1/7 (negative reciprocal of 7).
Now we have the slope of the normal line (-1/7) and a point on the curve (let's call it P(x₁, y₁)) to find the equation.
The general equation of a line in point-slope form is y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line and m is its slope.
Since we don't have any specific point, we'll use the equation of the curve itself.
To find the equation of the normal, substitute the slope (-1/7) and the general point (x, y) into the point-slope form:
y - (2x^2 + 3x + 4) = (-1/7)(x - x₁)
Expand and simplify the equation:
y - 2x^2 - 3x - 4 = (-1/7)x + (1/7)x₁
Move all terms to one side to get the equation in the standard form:
2x^2 + (1/7)x + y - (1/7)x₁ - 3x - 4 = 0
Finally, rewrite the equation in the standard form:
2x^2 - (20/7)x + y - (1/7)x₁ - 4 = 0
Therefore, the equation of the normal to the curve y=2x^2+3x+4 which is perpendicular to the line y=7x-5 is:
2x^2 - (20/7)x + y - (1/7)x₁ - 4 = 0