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Focal point of the parabola y = x² [Series N: Naked Truth] 
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Olinguito

1215778. Fri Dec 09, 2016 6:24 pm 


The focal point of the parabola y = f(x) = x² is the point (0,1/4). All beams of light shining vertically from the top will, after reflecting from the curve, pass through this point.
Suppose such a beam of light hits the parabola at the point P with coordinates (c,c²) where c > 0 (the case c < 0 is symmetrically identical). The tangent to the parabola at this point makes an angle θ with the horizontal, where tanθ = f'(c) = 2c. This is also the angle of incidence of the light beam (angle between incident beam and normal to the parabola at the point of incidence). The reflected beam makes an angle 2θ with the incident beam and so the (obtuse or reflex) angle between the reflected beam and the horizontal is 90°+2θ. Thus the slope of the reflected beam is
tan(90°+2θ)
= −cot(2θ)
= −(1 − tan²θ)/2tanθ
= −(1 − (2c)²)/2(2c)
= (4c² − 1)/4c
As the reflected beam passes through the point P, its Cartesian equation is
y − c² = [(4c² − 1)/4c](x − c)
When x = 0,
y − c² = [(4c² − 1)/4c](−c)
giving y = 1/4, a value independent of c. 




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