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MANOJ KUMAR Grade: 12
        

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

6 years ago

Answers : (2)

Aman Bansal
592 Points
										

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


 

6 years ago
Ashhad Quadri
24 Points
										
  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.
 
2 years ago
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