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Question

# Let the curve y=f(x) pass through the origin. If the mid point of the line segment of its normal between any point on the curve and the x−axis lies on the parabola y2=4x, then the equation of the curve y=f(x) satisfies

A
y2=16x+6464ex/4
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B
y2=16x+64+64ex/4
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C
y2=16x6464ex/4
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D
y2=16x64+64ex/4
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Solution

## The correct option is A y2=16x+64−64ex/4Let (x,y) be a point on the curve y=f(x) and (a,0) be a point on the x−axis. Slope of the normal is −dxdy=y−0x−a⇒x−a=−ydydx⇒a=x+ydydx ⋯(1) Now, the mid-point of the line segment of the normal is (x+a2,y2) Putting the mid-point in the given equation of the parabola, y24=4(x+a2)⇒a=y28−x ⋯(2) From equation (1) and (2), x+ydydx=y28−x⇒2ydydx−y24=−4x Assuming y2=t⇒2ydydx=dtdx So, the differential equation becomes dtdx−t4=−4x I.F.=exp(∫−14 dx)=e−x/4 So, the solution of the differential equation is, t⋅e−x/4=∫−4xe−x/4 dx+C where C is a constant of integration. ⇒t⋅e−x/4=−4[−4xe−x/4+4∫e−x/4 dx]+C⇒y2⋅e−x/4=−4[−4xe−x/4−16e−x/4]+C The curve passes through the origin. So putting (0,0), we get C=−64 Therefore, y2⋅e−x/4=−4[−4xe−x/4−16e−x/4]−64 ⇒y2=16x+64−64ex/4

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