# 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 $\frac{1}{P{K}^{2}}+\frac{1}{Q{K}^{2}}$ 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 ${y}^{2}=4ax$.
let the point K be (c,0)
Equation of line PQ using parametric coordinates: $x=c+r\mathrm{cos}\theta$ (Equation 1), $y=r\mathrm{sin}\theta$ (Equation 2)
From equation 1 and 2: ${\left(x-c\right)}^{2}+{y}^{2}={r}^{2}$ (Equation 3)
Using Equation 3 and ${y}^{2}=4ax$, we get this quadratic in r: ${r}^{2}-4ax-{\left(x-k\right)}^{2}=0$ (Equation 4)
Roots of this quadratic would be the lengths of PK and QK
From Equation 4, ${r}_{1}+{r}_{2}=0$ and ${r}_{1}{r}_{2}=-\left(4ax+{\left(x-k\right)}^{2}\right)$
We know, $\frac{1}{P{K}^{2}}+\frac{1}{Q{K}^{2}}=\frac{1}{{r}_{1}^{2}}+\frac{1}{{r}_{2}^{2}}=\frac{{\left({r}_{1}+{r}_{2}\right)}^{2}-2{r}_{1}{r}_{2}}{{\left({r}_{1}{r}_{2}\right)}^{2}}=\frac{-2}{{r}_{1}{r}_{2}}$
As value of ${r}_{1}{r}_{2}$ is not constant thus $\frac{1}{P{K}^{2}}+\frac{1}{Q{K}^{2}}$ 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?
You can still ask an expert for help

Expert Community at Your Service

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Available 24/7
• Math expert for every subject
• Pay only if we can solve it

## Answers (1)

Lilliana Rich
Answered 2022-03-10 Author has 3 answers
Roots of this quadratic 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: $x=c+r\mathrm{cos}\theta$
${y}^{2}=4ax⇒{r}^{2}{\mathrm{sin}}^{2}\theta -4ar\mathrm{cos}\theta -4ac=0$
Absolute values of roots of this equation would be the PK and QK.
${r}_{1}+{r}_{2}=\frac{4a\mathrm{cos}\theta }{{\mathrm{sin}}^{2}\theta }$
$\frac{1}{P{K}^{2}}+\frac{1}{Q{K}^{2}}=\frac{{\left({r}_{1}+{r}_{2}\right)}^{2}-2{r}_{1}{r}_{2}}{{\left({r}_{1}{r}_{2}\right)}^{2}}=\frac{16{a}^{2}{\mathrm{cos}}^{2}\theta +8ac{\mathrm{sin}}^{2}\theta }{16{a}^{2}{c}^{2}}$
This value does not depend on . This is the answer.
###### Not exactly what you’re looking for?

Expert Community at Your Service

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Available 24/7
• Math expert for every subject
• Pay only if we can solve it