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Grade: 11
        
 

Length of the chord

As in the preceding article, the abscissae of the points common to the straight line y = mx + c and the parabola y2 = 4ax are given by the equation m2x2 + (2mx – 4a) x + c2 = 0. 

={[4(mc-2a)^2]/m^4}-{(4c^2)/m^2}...........HOW????

2 years ago

Answers : (1)

Vijaya
11 Points
							
To obtain the points common to the straight line y = mx + c and the parabola y2= 4ax :
  1. Substitue y= mx + c in the given equation of parabola y2= 4ax.
  2. Doing that we get (mx +c)2= 4ax.
    > m2x2 + c2 + 2mxc = 4ax 
    > m2x+ c2 + 2mxc -4ax =0
    > m2x2 + 2x (mc- 4a) + c2 = 0
     the above equation is quadratic equation in x
     to obtain the abcissae of the points in common to the straight line y= mx+c and the parabola y2 = 4ax use the quadratic             roots formula  
     dividing  the above equation with m2
    > x2+ 2x(mc – 2a)/m2 + c2/ m2 = 0 …...(1)
    calculate the discriminant for the above eqaution 
    D= [2(mc – 2a)/ m2]2 – 4(c2 / m2)
       = {4(mc – 2a)2 /m4 } – {4(c2m2)}
 
2 years ago
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