Find the equation of normal to the curve 3x^2 - y^2 =8 which is parallel to the line x + 3y =4.
To find the equation of the normal to a curve that is parallel to a given line, we need to follow these steps:
Step 1: Find the derivative of the curve.
Step 2: Calculate the slope of the given line.
Step 3: Use the negative reciprocal of the slope of the curve as the slope of the normal.
Step 4: Use the point of tangency on the curve to find the y-intercept of the normal.
Step 5: Write the equation of the normal using the slope-intercept form.
Let's go through these steps to find the equation of the normal to the curve 3x^2 - y^2 = 8 which is parallel to the line x + 3y = 4.
Step 1: Find the derivative of the curve.
Differentiating the given equation with respect to x, we get:
6x - 2y(dy/dx) = 0
Step 2: Calculate the slope of the given line.
Rewrite the equation of the line in the slope-intercept form (y = mx + c):
3y = -x + 4
y = (-1/3)x + 4/3
Comparing this with y = mx + c, we see that the slope (m) of the line is -1/3.
Step 3: Use the negative reciprocal of the slope of the curve as the slope of the normal.
In the equation obtained from differentiating the curve (6x - 2y(dy/dx) = 0), the term dy/dx represents the slope of the curve. Therefore, the slope of the curve is dy/dx = 6x / 2y = 3x / y.
The negative reciprocal of the slope of the curve is -y / 3x.
Step 4: Use the point of tangency on the curve to find the y-intercept of the normal.
To find the point of tangency on the curve, we need to solve the curve equation simultaneously with the line equation.
Solving the equations:
3x^2 - y^2 = 8
x + 3y = 4
We get the point of tangency as (1, 1).
Now, substitute the point (1, 1) into the equation of the normal to find the y-intercept. Let's call it b.
Using the point-slope form, we have:
y - 1 = (-y / 3x)(x - 1)
y - 1 = (-y / 3 + 1/3)
3y - 3 = -y + 1
4y = 4
y = 1
Therefore, the y-intercept (b) of the normal is 1.
Step 5: Write the equation of the normal using the slope-intercept form.
Using the point-slope form with the slope (-y / 3x) and y-intercept of 1, we have:
y - 1 = (-y / 3x)(x - 1)
To simplify, let's multiply through by 3x:
3xy - 3x = -y(x - 1)
3xy - 3x = -xy + y
4xy = 4x + 3x
4xy = 7x
Dividing both sides by x:
4y = 7
Finally, rearrange the equation to the slope-intercept form:
y = 7/4
Therefore, the equation of the normal to the curve 3x^2 - y^2 = 8, which is parallel to the line x + 3y = 4, is y = 7/4.