If the maximum and minimum values of $|\vec{PA}||\vec{PB}|$ are $M$ and $m$ respectively then prove that the value of $M^2+m^2=34$

Question

Given three points on the $xy$ plane on $O(0,0),A(1,0)$ and $B(-1,0)$.Point $P$ is moving on the plane satisfying the condition $(\vec{PA}.\vec{PB})+3(\vec{OA}.\vec{OB})=0$
If the maximum and minimum values of $|\vec{PA}||\vec{PB}|$ are $M$ and $m$ respectively then prove that the value of $M^2+m^2=34$

---

My Attempt:
Let the position vector of $P$ be $x\hat{i}+y\hat{j}$.Then $\vec{PA}=(1-x)\hat{i}-y\hat{j}$ and $\vec{PB}=(-1-x)\hat{i}-y\hat{j}$
$\vec{OA}=\hat{i},\vec{OB}=-\hat{i}$
$(\vec{PA}.\vec{PB})+3(\vec{OA}.\vec{OB})=0$ gives
$x^2-1+y^2-3=0$
$x^2+y^2=4$
$|\vec{PA}||\vec{PB}|=\sqrt{(1-x)^2+y^2}\sqrt{(-1-x)^2+y^2}=\sqrt{(x^2+y^2-2x+1)(x^2+y^2+2x+1)}$
$|\vec{PA}||\vec{PB}|=\sqrt{(5-2x)(5+2x)}=\sqrt{25-4x^2}$
I found the domain of $\sqrt{25-4x^2}$ which is $\frac{-5}{2}\leq x \leq \frac{5}{2}$.Then i found the minimum and maximum values of $\sqrt{25-4x^2}$ which comes out to be $M=5$ and $m=0$.So $M^2+m^2=25$

But i need to prove $M^2+m^2=34$.Where have i gone wrong?Please help me.Thanks.

Answer 1

You have almost Got it. Here $x^2+y^2 = 4\;,$ Then Parametric Coordinats are

$x=2\cos \theta\;\;,y=2\sin \theta$ and Here $f(x) = \sqrt{25-4x^2}.$

So we get $f(\theta) = \sqrt{25-16\cos^2 \theta}\;,$ Now Using $0\leq \cos^2 \theta \leq 1$

So $\displaystyle \min\left[f(\theta)\right] = \sqrt{25-16} = 3$ at $\displaystyle \theta =0,2\pi.$

and $\displaystyle \max\left[f(\theta)\right] = \sqrt{25-0} = 5$ at $\displaystyle \theta =\frac{\pi}{2}\;\ \frac{3\pi}{2}.$

Answer 2

You are almost there. However, the limit you set for $x$ is wrong. Please note that: $$x^2 + y^2 = 4 \to x^2 \le 4 \to -x^2 \ge -4$$ Then, putting into your equation of PA.PB gives: $$PA.PB \ge \sqrt{25 - 4\times 4} = 3$$ Thus, m = 3 and M = 5 which yields $$m^2 + M^2 = 9 + 25 = 34$$