# THREE NORMALS ARE DRAWN FROM THE POINTS (14,7) TO THE CURVE y2-16X-8Y=0. FIND THE COORDINATES OF FEET OF NORMALS.NOTE. IN EQUNTION 2 IS POWER OF y

Aman Bansal
592 Points
13 years ago

dear manoj,

let the coordinates of foot of perprndicular be (x1,y1)

differentiating the equation to find the slope of the normal,

=> dy/dx = 8/(y-4)

slope = -(dx/dy)

hence we can find the equation of normal in terms of x1 and y1.

then passing through the (x1,y1)

thus we find a cubic equation on solving which we get,

y1= 0, 16 , -4

thus corresponding x1 = 0, 8 , 3

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AMAN BANSAL

24 Points
8 years ago
1. Y=mx-2am-am3
As we know that this equation implies for standard  parabola
But the given equation  is in shifted form
After simplifying  the equation
(Y-4 )2=16 (X+1)
Replacing y-4 =Y
And x+1=1
Placing in equation of normal and x=14 and y=7 we get m in cubic find the roots
Remember point of contact also changes.