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Question

If the tangent to the curve, y=ex at a point c,ec and the normal to the parabola,y2=4x at the point 1,2 intersect at the same point on the x-axis, then the value of c is


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Solution

Step 1: Finding the relation between x and c:

y=ex

Differentiate this with respect to x to get the slope,

dydx=ex⇒m=dydxc,ec=ec

Equation of tangent at point c,ec,

y−ec=ecx−c[∵y-y1=m(x-x1)]

It crosses the x-axis

So, substitute y=0 in the above equation:

x=c−1...i

Step 2: Finding the value of x:

Now,

y2=4x

Differentiate the above equation with respect to x to get the slope,

2ydydx=4⇒m=dydx1.2=22=1

The slope of normal=-1m.

=-1

The equation of normal is

y−2=−1x−1[∵y-y1=mx-x1]⇒x+y=3

It crosses the x axis .So, substitute y=0 in the above equation,

x+0=3⇒x=3

Step 3: Find the value of c:

By substituting the value of x in the equation i as tangent and normal meet at same point, we get

3=c-1⇒c=4

Hence, the value of c is 4.


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