Equation of Tangent at a Point (x,y) in Terms of f'(x)
The normal to...
Question
The normal to the curve x2=4y passing (1,2) is
A
x+y=3
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B
x−y=3
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C
x+y=1
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D
x−y=1
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E
answer required
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Solution
The correct option is Ex+y=3 Given, x2=4y
Slope of tangent to the curve will be 4dydx=2x ⇒dydx=x2 So slope of normal will be −dxdy=−2x Let the point on the curve be (h,k) So slope of normal will be −2h Equation of normal is given by (y−k)=−2h(x−h) It is given that the normal passes through (1,2) ⇒(2−k)=−2h(1−h).............(1) Also, (h,k) lies on the curve so h2=4k From (1) we have h34=2h+2−2h=2 ⇒h3=8 ⇒h=2 k=h24⇒k=1 So the equation will be y−1=−22(x−2) x+y=3