A straight line L with negative slope passes through the point (8,2) and cuts the positive coordinate...

Question

Problem :

A straight line L with negative slope passes through the point (8,2) and cuts the positive coordinate axes on points P and Q . As L varies find the absolute minimum value of OP + OQ ( O is origin ) .

My approach :

Let us assume points P and Q as P(0, a sin$\theta) , Q ( acos\theta, 0)$ and after that can we find the distance of OP and OQ , but this wont work as I am not getting value of $\theta$ Please suggest how to proceed thanks.

Answer 1

Let the x and y-intercept be $a$ and $b$ respectively.

Equation of line :

$L:\frac{x}{a}+\frac{y}{b}=1$

It passes through $S(8,2)$

$\frac{8}{a}+\frac{2}{b}=1$

$b=\frac{2a}{a-8}$

Let $z=a+b$

$z=a+\frac{2a}{a-8}$

$=\frac{a^2-6a}{a-8}$

Now find minimum value of $z$ by derivative technique.

Can you solve it from here?

Answer 2

My Solution:: Let the equation of the line be $\displaystyle \frac{x}{a}+\frac{y}{b}=1$. Given that the line passes through the point $(8,2)$ and its slope is negative, that means $a,b>0$, and put $x=8,y=2$ in $$\frac{x}{a}+\frac{y}{b}=1$$

We get $$\frac{8}{a}+\frac{2}{b} = 1.$$

Now using the Cauchy-Schwarz inequality ......

$$\Rightarrow \frac{\left(2\sqrt{2}\right)^2}{a}+\frac{\left(\sqrt{2}\right)^2}{b}\geq \frac{(2\sqrt{2}+\sqrt{2})^2}{a+b} = \frac{18}{a+b}$$.

So we get $$a+b\geq 18,$$ Where $a=OA.$ and $b=OB$.