How do you find the equation of the normal line to the parabola #y=x^25x+4# that is parallel to the line #x3y=5#?
The normal is:
The general equation of the normal line is:
If we put the equation of the line in the same form:
we can see the two lines are parallel when
Take the derivative of f(x):
and fond the value of
The desired normal line is:
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To find the equation of the normal line to the parabola y=x^25x+4 that is parallel to the line x3y=5, we need to determine the slope of the normal line.
The slope of the given line can be found by rearranging it into slopeintercept form (y = mx + b), which gives us 3y = x  5. Dividing both sides by 3, we get y = (1/3)x  5/3.
Since the normal line is parallel to this line, it will have the same slope. Therefore, the slope of the normal line is also 1/3.
To find the point on the parabola where the normal line intersects, we need to find the derivative of the parabola equation. Taking the derivative of y=x^25x+4 with respect to x gives us dy/dx = 2x  5.
Next, we set the derivative equal to the negative reciprocal of the slope of the normal line (since the normal line is perpendicular to the tangent line). So, 2x  5 = 1/3.
Solving this equation, we find x = 4.
Substituting this value of x back into the original equation of the parabola, we get y = (4)^2  5(4) + 4 = 4.
Therefore, the point of intersection is (4, 4).
Using the pointslope form of a line (y  y1 = m(x  x1)), where m is the slope and (x1, y1) is a point on the line, we can write the equation of the normal line as y  4 = (1/3)(x  4).
Simplifying this equation, we get y = (1/3)x + 8/3.
Hence, the equation of the normal line to the parabola y=x^25x+4 that is parallel to the line x3y=5 is y = (1/3)x + 8/3.
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To find the equation of the normal line to the parabola (y = x^2  5x + 4) that is parallel to the line (x  3y = 5), first, find the derivative of the parabola to get the slope of the tangent line at any point. Then, use the fact that the normal line is perpendicular to the tangent line, so its slope will be the negative reciprocal of the slope of the tangent line. Once you have the slope of the normal line, use a point on the parabola to find the equation of the normal line using the pointslope form.

Find the derivative of the parabola: (y = x^2  5x + 4) (y' = 2x  5)

Find the slope of the tangent line at any point on the parabola: For the parabola, the slope of the tangent line at any point is given by the derivative: (m_{\text{tangent}} = 2x  5)

Find the slope of the normal line: Since the normal line is parallel to the given line, (x  3y = 5), its slope will be the same as the slope of that line. So, rearrange the equation to (y = \frac{1}{3}x  \frac{5}{3}), and the slope is (m_{\text{normal}} = \frac{1}{3}).

Use the fact that the product of slopes of perpendicular lines is 1 to find the slope of the normal line: (m_{\text{tangent}} \cdot m_{\text{normal}} = 1) ((2x  5) \cdot \frac{1}{3} = 1) (2x  5 = 3) (2x = 2) (x = 1)

Find the corresponding yvalue using the equation of the parabola: (y = 1^2  5(1) + 4) (y = 1  5 + 4) (y = 0)

Use the pointslope form of the equation of a line to find the equation of the normal line: (y  y_1 = m(x  x_1)) (y  0 = \frac{1}{3}(x  1)) (y = \frac{1}{3}x  \frac{1}{3})
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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