Question

The lines joining the origin to the point of intersection of $3x^2+\lambda xy-4x+1=0and2x+y-1=0$ are at right angles for

• A. $\lambda =-4$
• B. $\lambda =4$
• C. $\lambda =7$
• D. no value of $\lambda$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\lambda =-4$
B $\lambda =4$
C $\lambda =7$

$3x^2+\lambda xy-4x+1=0$

$\Rightarrow 3x^2+\lambda xy-4x\left(1\right)+(1{)}^2=0$

$2x+y-1=0$

$3x^2+\lambda xy-4x(2x+y)+(2x+y{)}^2=0$

$3x^2+\lambda xy-8x^2+4xy+4x^2+y^2+4xy=0$

$\Rightarrow -x^2+xy(m+4)+y^2=0$

$\Rightarrow x^2-xy(m+4)-y^2=0$

$\tan \alpha =\frac{2\sqrt{b^2-ab}}{a+b}$

$a=1;b=-1$

$\tan \alpha =\frac{2\sqrt{1-1}}{1+1}$

$\tan \alpha =\infty$

$\infty =\tan \alpha$

$\Rightarrow \alpha ={\tan }^{-1}\left(\infty \right)={90}^o$

$\therefore \lambda$ can be any value from the options

$\therefore \lambda =-4;\lambda =4;\lambda =7$.