$Q \equiv\left( x _{2}, mx _{2}\right)$
$A _{1}=\frac{1}{2}\left|\begin{array}{ccc}3 & 4 & 1 \\ 2 & 0 & 1 \\ -1 & 1 & 1\end{array}\right|=\frac{13}{2}$
$A _{2}=\frac{1}{2}\left|\begin{array}{ccc} x _{1} & mx _{1} & 1 \\ x _{2} & mx _{2} & 1 \\ 2 & 0 & 1\end{array}\right|$
$A _{2}=\frac{1}{2}\left|2\left( mx _{1}- mx _{2}\right)\right|= m \left| x _{1}- x _{2}\right|$
$A _{1}=3 A _{2} \Rightarrow \frac{13}{2}=3 m \left| x _{1}- x _{2}\right|$
$AC : x +3 y =2$
$BC : y =4 x -8$
$P : x +3 y =2$ and $y = mx \Rightarrow x _{1}=\frac{2}{1+3 m }$
$Q : y =4 x -8$ and $y = mx \Rightarrow x _{2}=\frac{8}{4- m }$
$\left| x _{1}- x _{2}\right|=\left|\frac{2}{1+3 m }-\frac{8}{4- m }\right|$
$=\left|\frac{-26 m }{(1+3 m )(4- m )}\right|=\frac{26 m }{(3 m +1)| m -4|}$
$=\frac{26 m }{(3 m +1)(4- m )}$
$\left| x _{1}- x _{2}\right|=\frac{13}{6 m }$
$\Rightarrow \frac{26 m }{(3 m +1)(4- m )}=\frac{13}{6 m }$
$\Rightarrow \quad 12 m ^{2}=-(3 m +1)( m -4)$
$\Rightarrow \quad 12 m ^{2}=-\left(3 m ^{2}-11 m -4\right)$
$\Rightarrow \quad 15 m ^{2}-11 m -4=0$
$\Rightarrow \quad 15 m ^{2}-15 m +4 m -4=0$
$\Rightarrow \quad(15 m +4)( m -1)=0$
$\Rightarrow m =1$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| $column 1$ | $column 2$ | $column 3$ |
| $(I)$ $x^2+y^2=a^2$ | $(i)$ $m y=m^2 x+a$ | $(P)$ $\left(\frac{a}{m^2}, \frac{2 a}{m}\right)$ |
| $(II)$ $x^2+a^2 y^2=a^2$ | $(ii)$ $y=m x+a \sqrt{m^2+1}$ | $(Q)$ $\quad\left(\frac{-m a}{\sqrt{m^2+1}}, \frac{a}{\sqrt{m^2+1}}\right)$ |
| $(III)$ $y^2=4 a x$ | $(iii)$ $y=m x+\sqrt{a^2 m^2-1}$ | $(R)$ $\quad\left(\frac{-a^2 m}{\sqrt{a^2 m^2+1}}, \frac{1}{\sqrt{a^2 m^2+1}}\right)$ |
| $(IV)$ $x^2-a^2 y^2=a^2$ | $(iv)$ $y=m x+\sqrt{a^2 m^2+1}$ | $(S)$ $\quad\left(\frac{-a^2 m}{\sqrt{a^2 m^2-1}}, \frac{-1}{\sqrt{a^2 m^2-1}}\right)$ |
($1$) The tangent to a suitable conic (Column $1$) at $\left(\sqrt{3}, \frac{1}{2}\right)$ is found to be $\sqrt{3} x+2 y=4$, then which of the following options is the only CORRECT combination?
$[A] (II) (iii) (R)$ $[B] (IV) (iv) (S)$ $[C] (IV) (iii) (S)$ $[D] (II) (iv) (R)$
($2$) If a tangent to a suitable conic (Column $1$) is found to be $y=x+8$ and its point of contact is $(8,16$ ), then which of the following options is the only CORRECT combination?
$[A] (III) (i) (P)$ $[B] (III) (ii) (Q)$ $[C] (II) (iv) (R)$ $[D] (I) (ii) (Q)$
($3$) For $a=\sqrt{2}$, if a tangent is drawn to a suitable conic (Column $1$ ) at the point of contact $(-1,1)$, then which of the following options is the only CORRECT combination for obtaining its equation?
$[A] (II) (ii) (Q)$ $[B] (III) (i) (P)$ $[\mathrm{C}]$ $(I) (1) (P)$ $[D] (I) (ii) (Q)$