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If a line y= mx+c cuts interest between the parabola y2=4ax.Then what is the length of interest in the terms of m, a, and c ?
let line L intersect parabola P at A(at1^2, 2at1) and B(at2^2, 2at2).clearly t1 and t2 both satisfy 2at= m*at^2+c.or mat^2 – 2at + c=0hence t1+t2= -(-2a)/ma= 2/m and t1.t2= c/maso length^2= (at1^2 – at2^2)^2 + (2at1 – 2at2)^2= a^2(t1 – t2)^2[(t1+t2)^2 + 4]= a^2[(t1+t2)^2 – 4t1t2][(t1+t2)^2 + 4]= a^2(s^2 – 4p)(s^2 + 4) where s= t1+t2 and p= t1.t2so length^2= a^2(4/m^2 – 4c/ma)(4/m^2 + 4)= 16(a/m^4)(a – mc)(1+m^2)hence length= (4/m^2)*sqrt(a(1+m^2)(a – mc))KINDLY APPROVE :))
Dear student For a line to be normal to the parabola at any point (x1,y1), it has to be normal to the tangent at that point as well, nowSlope(derivative w.r. t x) of tangent at any point on parabola (x1,y1)m = 4a/(2*y1)If slope of tangent is m, then the slope of normal will bem'= (-1/m)Hence, any line having slope m' and passing through the same point (x1,y1) will be a normal to the parabola.Hope this answers your question. :)Good Luck
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