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Question

The area of the triangle formed by the tan-gent and the normal at (a,a) on the curve y2=x32ax and the line x=2a is (in sq.units)

A
7a24
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B
a22
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C
5a24
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D
9a24
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Solution

The correct option is C 5a24
Equation of the given curve is y2(2ax)=x3

Differentiating both sides w.r.t x, we get

y2×(1)+(2ax)×2ydydx=3x2

(2ax)2ydydx=3x2+y2

dydx=3x2+y22y(2ax)

Slope of tangent at (a,a) is m1=3a2+a22a(2aa)=4a22a2=2

Slope of normal at (a,a) is m2=12

Equation of tangent at (a,a) is

ya=2(xa)

y=2x2a+a=2xa ......(1)

Equation of normal at (a,a) is

ya=12(xa)

2y=x+a+2a=x+3a

y=x2+3a2 ......(2)

Equation of the given line is

x=2a ...(3)

Solving (1) and (3), we have
x=2a and y=2(2a)a=4aa=3a

Solving (2) and (3), we have
x=2a and y=2a2+3a2=a2

Solving (1) and (2), we have
x=a and y=a

Let A(2a,3a),B(2a,a2) and C(a,a) be the vertices of the triangle.

Area of ABC=122a(a2a)+2a(a3a)+a(3aa2)

=122a(a2)+2a(2a)+a(5a2)

=12a24a2+5a22

=1210a2+5a22

=5a24

Thus, the area of triangle formed is 5a24 square units.

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