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+rcos⁡θ (Equation 1), y=rsin⁡θ (Equation 2)From equation 1 and 2: (x−c)2+y2=r2 (Equation 3)Using Equation 3 and y2=4ax, we get this quadratic in r: r2−4ax−(x−k)2=0 (Equation 4)Roots of this quadratic (r1 and r2) would be the lengths of PK and QKFrom Equation 4, r1+r2=0 and r1r2=−(4ax+(x−k)2)We know, 1PK2+1QK2=1r12+1r22=(r1+r2)2−2r1r2(r1r2)2=−2r1r2As value of r1r2 is not constant thus 1PK2+1QK2 does not turn out to be constant. Hence, this solution is incorrect.I've seen the correct solution but I wanted to know why this solution is incorrect?

Question

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+rcos⁡θ (Equation 1), y=rsin⁡θ (Equation 2)From equation 1 and 2: (x−c)2+y2=r2 (Equation 3)Using Equation 3 and y2=4ax, we get this quadratic in r: r2−4ax−(x−k)2=0 (Equation 4)Roots of this quadratic (r1 and r2) would be the lengths of PK and QKFrom Equation 4, r1+r2=0 and r1r2=−(4ax+(x−k)2)We know, 1PK2+1QK2=1r12+1r22=(r1+r2)2−2r1r2(r1r2)2=−2r1r2As value of r1r2 is not constant thus 1PK2+1QK2 does not turn out to be constant. Hence, this solution is incorrect.I&#039;ve seen the correct solution but I wanted to know why this solution is incorrect?

Lilliana Rich · Accepted Answer

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+rcos⁡θy2=4ax⇒r2sin2θ−4arcos⁡θ−4ac=0Absolute values of roots of this equation would be the PK and QK.r1+r2=4acos⁡θsin2θ1PK2+1QK2=(r1+r2)2−2r1r2(r1r2)2=16a2cos2θ+8acsin2θ16a2c2This value does not depend on θ at c=2a. This is the answer.

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

Answered question

Answer & Explanation

New Questions in Integral Calculus