Question

Consider a family of lines $(4a+3)x-(a+1)y-(2a+1)=0$ where $a\in R$
Minimum area of the triangle which a member of this family with negative gradient can make with the positive semi axes, is

• A. 8
• B. 6
• C. 4
• D. 2

Answer 1

Answer: (C)

Given $(4a+3)x-(a+1)y-(2a+1)=0$
$\Rightarrow (3x-y-1)+a(4x-y-2)=0$
family of lines passes through the fixed point $P$ which is the intersection of $3x-y=1$ and $4x-y=2$
$\therefore$ Solving $P(1,2)$
(i) $\frac{k}{h}\cdot \frac{k-2}{h-1}=-1$

$\therefore$ Locus is $x(x-1)+y(y-2)=0$
$\Rightarrow x^2+y^2-2y-x=0$
$\Rightarrow {\left(x-\frac{1}{2}\right)}^2+(y-1{)}^2=\frac{5}{4}$
$\Rightarrow (2x-1{)}^2+4(y-1{)}^2=5$
(ii) We have $y-2=m(x-1)$
This makes an angle of $\pi /4$ with $3x-4y=2$ with slope $3/4$.
$\therefore \left|\frac{m-(3/4)}{1+(3m/4)}\right|=1;\frac{4m-3}{4+3m}=\pm 1$
$\Rightarrow 4m-3=4+3m$ (with positive sign) $;m=7$
with negative sign ; $4m-3=-4-3m$
$\Rightarrow 7m=-1$
$\Rightarrow m=\frac{-1}{7}$ (rejected)
Hence the line is $y-2=7(x-1);7x-y-5=0$
(iii) Again $(y-2)=m(x-1)$
$x=0;y=2-m;y=0,x=1-\frac{2}{m}$
$\therefore 2A=(2-m)\left(1-\frac{2}{m}\right)(m<0)$
$\Rightarrow 2A=2-m-\frac{4}{m}+2=4+\left(-m-\frac{4}{m}\right)$
Let $-m=M(M>0);2A=4+M+\frac{4}{M}$
$=4+{\left(\sqrt{M}-\frac{2}{\sqrt{M}}\right)}^2+4=8+{\left(\sqrt{M}-\frac{2}{\sqrt{M}}\right)}^2$
Area is minimum if $M=2$
$\Rightarrow m=-2$
${2A|}_{\min }=8$
$\Rightarrow A{|}_{\min }=4$