Author Topic: Parametrics Question (Read 1434 times) Tweet Share

frog0101 · « **on:** October 31, 2017, 02:13:28 pm »

Hi,
I am unable to prove this question out of the Cambridge textbook, any help would be great:
P is a variable point on the parabola x^2=4y. The normal at P meets the parabola again at Q. The tangents at P and Q meet at T. S is the focus and QS=2PS.
Prove that angle (PSQ)=pi/2

Thanks

RuiAce · « **Reply #1 on:** October 31, 2017, 05:33:21 pm »

$\text{For the variable point }P(2p,p^2)\\ \text{it can be shown that the normal at }P\text{ to }x^2=4y\text{ is}\\ \boxed{x+py=2p+p^3}$
$\text{To find the point }Q\text{, we intersect this line again with }\boxed{x^2=4y}\\ \text{This gives }x + \frac{px^2}{4} = 2p+p^3$
$\text{We may use the quadratic formula to find the other root, i.e. the}\\ x\text{-coordinate of }Q.\\ \text{Alternatively, because we know }x=2p\text{ is one of the roots}\\ \text{the sum of roots gives }2p + \beta = -\frac4p \implies \boxed{\beta = -\frac4p-2p}$
$\text{Hence, }Q\text{ is the point}\\ \boxed{Q\left(-\frac2p (2+p^2), \frac{(2+p^2)^2}{p^2} \right)}$
______________________________
$\text{Now, }S\text{ is the point }(0,1).\\ \text{Essentially we want to prove that }\triangle PSQ\text{ is right at }S\\ \text{so we need to prove that }PQ^2 =<br />PS^2+QS^2$
$\text{Because }QS = 2PS\\ \text{we've been given sufficient information to find our value for }p.\\ \text{Applying the distance formula in }QS^2 = 4PS^2\text{ gives}\\ \boxed{\frac{4(p^2+2)^2}{p^2} + \left(\frac{(p^2+2)^2}{p^2}-1\right)^2 = 4\left[(2p)^2 + (p^2-1)^2\right]}$
$\text{The LHS rearranges to give}\\ \begin{align*}\frac{4(p^2+2)^2}{p^2} + \left(\frac{(p^2+2)^2}{p^2}-1\right)&= \frac{(4p^2+2)^2}{p^2} + \left[\frac{(p^2+2)^2}{p^2}\right]^2 - \frac{2(p^2+2)^2}{p^2} + 1\\ &= \left[\frac{(p^2+2)^2}{p^2}\right]^2 + \frac{2(p^2+2)^2}{p^2} + 1 \\ &= \left(\frac{(p^2+2)^2}{p^2}+1\right)^2\end{align*}$
$\text{Similarly, the RHS rearranges to}\\ 4\left[(2p)^2 + (p^2-1)^2\right] = 4(p^2+1)^2$
$\text{So our equation is now}\\ \boxed{\left(\frac{(p^2+2)^2}{p^2}+1\right)^2 = 4(p^2+1)^2}$
______________________________

Clearly, taking negative roots will give us a contradiction, as one side will be positive whilst the other is negative.
$\text{So our equation now reduces to}\\\frac{(p^2+2)^2}{p^2}+1 =2(p^2+1)\\ \text{This is now easy to solve.}$
\begin{align*}(p^2+2)^2 + p^2 &= 2p^2(p^2+1)\\ p^4+4p^2 + 4 + p^2 &= 4p^4 + 2p^2\\ p^4 - 3p^2 - 4 &= 0\\ (p^2-4)(p^2+1)&=0\\ (p+2)(p-2)(p^2+1)&=0\\ p&=\pm 2\end{align*}
$\text{Due to the symmetry of the parabola}\\ \text{it suffices to only take the positive case }\boxed{p=2.}$
______________________________
$\text{So we now know that }P\text{ is }(4,4)\\ \text{and }Q\text{ is }(-6,9)\\ \text{The distance formula can easily verify that }PQ^2+PS^2+QS^2.$

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Author Topic: Parametrics Question (Read 1434 times) Tweet Share

frog0101

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