Let $P(x,y)$ and point on Hyperbola $3x^2-4y^2=36$, which is nearest to line $3x+2y=1.$ Then $\displaystyle \sqrt{2}(y-x)=$
What I try:
First we will find perpendicular distance of line $3x+2y-1=0$ from point $P(x,y),$ which is $\displaystyle d=\bigg|\frac{3x+2y-1}{\sqrt{3^2+2^2}}\bigg| =\bigg|\frac{3x+2y-1}{\sqrt{13}}\bigg|$
Now i am trying using Titu,s leema
$\displaystyle \frac{(3x)^2}{108}+\frac{(2y)^2}{-36}\geq \frac{(3x+2y)^2}{108-36}$
$\displaystyle (3x+2y)^2\leq 72\Longrightarrow (3x+2y)\leq 6\sqrt{2}$
And equality hold, when $\displaystyle \frac{(3x)}{108}=-\frac{(2y)}{36}\Longrightarrow x=-2y$
Put into $3x^2-4y^2=36\Longrightarrow x=\pm \frac{3}{\sqrt{2}}$ and $\displaystyle y=\mp 3\sqrt{2}$
But Here is my doubt is when we put $3x+2y=6\sqrt{2}$ in $\displaystyle d\leq \frac{6\sqrt{2}-1}{\sqrt{13}}$
Which is maximum value of $d$, But here we have to obtain minimum of $d$
Please help me , Thanks
Best Answer
This is not correct.
Titu's lemma says that if $a_{1},a_{2},b_{1},b_{2}$ are real numbers with $\color{red}{b_1\gt 0,b_2\gt 0}$, then $$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}\ge\frac{(a_1+a_2)^2}{b_1+b_2}$$ So, you cannot use Titu's lemma here.
You can use the following inequality :
If $a_{1},a_{2},b_{1},b_2$ are real numbers with $\color{red}{0\gt b_2\gt -b_1}$, then $$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}\color{red}{\le}\frac{(a_1+a_2)^2}{b_1+b_2}\tag1$$The equality is attained only when $a_1b_2 - a_2b_1=0$.
Proof for $(1)$ : $$\begin{align}&\frac{(a_1+a_2)^2}{b_1+b_2}-\bigg(\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}\bigg)\\\\&=\frac{b_1b_2(a_1+a_2)^2-a_1^2b_2(b_1+b_2)-a_2^2b_1(b_1+b_2)}{b_1b_2(b_1+b_2)} \\\\&=\frac{-(a_1^2b_2^2-2a_1a_2b_1b_2+a_2^2b_1^2)}{b_1b_2(b_1+b_2)} \\\\&=\frac{(a_1b_2 - a_2b_1)^2}{b_1(-b_2) (b_1 + b_2)}\ge 0.\ \blacksquare\end{align}$$
Using $(1)$, we have $$ \frac{(3x)^2}{108}+\frac{(2y)^2}{-36}\color{red}{\le} \frac{(3x+2y)^2}{108-36}$$
So, we have $$(3x+2y)^2\color{red}{\ge} 72 \implies |3x+2y|\ge 6\sqrt{2}$$
I think you can continue from here.