Prove that the straight line y=mx+c touches the parabola y2 =4a (x+a)if c=am+a/m
Answers
Given : straight line y=mx+c touches the parabola y2 =4a (x+a)
To Find : Prove that c = am + a/m
Solution:
y² = 4a(x + a)
=> 2y dy/dx = 4a
=> dy/dx = 2a/y
dy/dx = Slope of Tangent ( touching line )
Let say point at which y=mx+c touches y² = 4a(x + a) is ( h , k)
=> dy/dx = m = 2a/k => k = 2a/m
y = mx + c
Substituting ( h , k)
=> k = mh + c
Substituting k = 2a/m
=> 2a/m = mh + c
=> h = (2a/m - c)/m
=> h = (2a - mc)/m²
y² = 4a(x + a)
Substituting ( h , k)
=> k² = 4a(h + a)
(2a/m)² = 4a((2a - mc)/m² + a)
=> 4a²/m² = 4a (2a - mc + am²)/m²
=> a = 2a - mc + am²
=> mc = a + am²
=> c = a/m + am
=> c = am + a/m
QED
Hence Proved
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