Question

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

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