Find a point on the curve y = (x − 2)2 at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).
Solution and Explanation
If a tangent is parallel to the chord joining the points (2, 0) and (4, 4), then the slope of the tangent = the slope of the chord.
The slope of the chord is \(\frac{4-0}{4-2}\) =\(\frac42\)=2.
Now, the slope of the tangent to the given curve at a point (x, y) is given by,
\(\frac{dy}{dx}\)=2(x-2)
Since the slope of the tangent = slope of the chord, we have:
2(x-2) = 2
x-2=1=x=3
when x=3,y=(3-2)2=1
Hence, the required point is (3, 1)
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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:
