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Question

What's the equation of a tangent on the parabola
y2=5x at the point (5, 5)


A

x + 2y + 5 = 0

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B

x - 5y + 2 = 0

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C

x - 2y + 5 = 0

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D

x + 5y - 2 = 0

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Solution

The correct option is C

x - 2y + 5 = 0


Method 1

Given equation of parabola is,

y2=5x .............(1)

The tangent will be having a slope equal to dydx(5,5)

and also passes through the point (5,5)

differentiating (1) with respect to x

2y.dydx=5

dydx=52y(5,5)=12

equation of the tangent

y5=12(x5)

2y-10=x-5

x-2y+5=0

Method 2

Here we can use the formula T1=0 for tangent

at the point (x1,y1).T1 is obtained by replacing

x2 with xx1

y2 with yy1

x with x+x12

y with y+y12

T1=0
yy1=5.[x+x12]

5y=5.(x+5)2

2y=x+5

x2y+5=0


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