Why is this solution incorrect? Prove that on the axis of any parabola there is a certain point

brecruicswbp 2022-03-09 Answered
Why is this solution incorrect?
Prove that on the axis of any parabola there is a certain point K which has the property that, if a chord PQ of the parabola be drawn through it, then 1PK2+1QK2 is the same for all positions of the chord.
If it would be valid for standard parabola than it would be valid for all parabolas. Thus, proving for y2=4ax.
let the point K be (c,0)
Equation of line PQ using parametric coordinates: x=c+rcosθ (Equation 1), y=rsinθ (Equation 2)
From equation 1 and 2: (xc)2+y2=r2 (Equation 3)
Using Equation 3 and y2=4ax, we get this quadratic in r: r24ax(xk)2=0 (Equation 4)
Roots of this quadratic (r1 and r2) would be the lengths of PK and QK
From Equation 4, r1+r2=0 and r1r2=(4ax+(xk)2)
We know, 1PK2+1QK2=1r12+1r22=(r1+r2)22r1r2(r1r2)2=2r1r2
As value of r1r2 is not constant thus 1PK2+1QK2 does not turn out to be constant. Hence, this solution is incorrect.
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Answers (1)

Lilliana Rich
Answered 2022-03-10 Author has 3 answers
Roots of this quadratic (r1 and r2) would be the lengths of PK and QK.
This part is incorrect. Roots of this equation correspond to the same value of x. P and Q can have different x.
Update: x=c+rcosθ
y2=4axr2sin2θ4arcosθ4ac=0
Absolute values of roots of this equation would be the PK and QK.
r1+r2=4acosθsin2θ
1PK2+1QK2=(r1+r2)22r1r2(r1r2)2=16a2cos2θ+8acsin2θ16a2c2
This value does not depend on θ at c=2a. This is the answer.
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