Find apoint on curve y=(x-2)^2 at which the tangent is parallel to the chord joining the points (2,0) n (4,4)?

Question

(x-2)^2 ---> (x-2) raised to the power 2 .

Please answer this question clearly with all the steps and do explain it to me.

This is from application of derivatives >tangents and normal.Also tell me what all properties are involved in this?

Answer 1

Ok. Given curve is a parabola y = (x - 2)^2, for its graph you can see http://farm3.static.flickr.com/2397/2784...

Point on the curve at which the tangent touches having some slope can be found out by differentiating the equation explained below in detail.

Step 1   (Differentiate the given equation)

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Eq. of given curve,

y = (x - 2)^2             ..................(i)

differentiating we get

dy/dx = 2 * (x - 2)      ..................(ii)

Step 2    (To find the X coordinate of the point)

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Now you have to find the point, right!, that means you have to find both x and y coordinates.

Now note that dy/dx can be calculated easily from the given data about the joining points of the chord.

As dy/dx is nothing but slope of the tangent to the curve at a fixed

point, (eq (ii) is giving us dy/dx as a function of x coordinate). Therefore dy/dx = slope of the tangent at point (x,y) on the curve = slope of the chord = (4 - 0) / (4 - 2) = 2

Now replace dy/dx in (ii) by its value 2. we get,

2 = 2 * (x - 2)

This gives us x = 3

Step 3    (To find the Y coordinate of the point)

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Now to find the y coordinate simply replace x by its value in (i). this gives us y = 1.

Therefore the point is (3, 1) at which the tangent touches the curve having slope = 2 = slope of the chord.