Question:
Find the equations of all lines having slope 0 which are tangent to the curve y=\(\frac{1}{x^2-2x+3}.\)
Find the equations of all lines having slope 0 which are tangent to the curve y=\(\frac{1}{x^2-2x+3}.\)
Updated On: Oct 13, 2023
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Solution and Explanation
The equation of the given curve is y=\(\frac{1}{x^2-2x+3}.\)
The slope of the tangent to the given curve at any point (x, y) is given by,
\(\frac{dy}{dx}\) =\(\frac{(-2x-2)}{(x^2-2x+3)^2}\) =\(\frac{-2(x-1)}{(x^2-2x+3)^2}\)
If the slope of the tangent is 0, then we have:
⇒ \(\frac{-2(x-1)}{(x^2-2x+3)^2}\)=0
⇒ -2(x-1)=0
⇒ x=1
When x = 1, y=\(\frac{1}{1-2+3}\) =\(\frac12\).
∴The equation of the tangent through(1,\(\frac12\)) is given by,
y-\(\frac12\)=0(x-1)
y-\(\frac12\)=0
y=\(\frac12\)
Hence, the equation of the required line is y=\(\frac12\)
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Concepts Used:
Tangents and Normals
- A tangent at a degree on the curve could be a straight line that touches the curve at that time and whose slope is up to the derivative of the curve at that point. From the definition, you'll be able to deduce the way to realize the equation of the tangent to the curve at any point.
- Given a function y = f(x), the equation of the tangent for this curve at x = x0
- Slope of tangent (at x=x0) m=dy/dx||x=x0
- A normal at a degree on the curve is a line that intersects the curve at that time and is perpendicular to the tangent at that point. If its slope is given by n, and also the slope of the tangent at that point or the value of the derivative at that point is given by m. then we got
m×n = -1
- The normal to a given curve y = f(x) at a point x = x0
- The slope ‘n’ of the normal: As the normal is perpendicular to the tangent, we have: n=-1/m
Diagram Explaining Tangents and Normal:
