Question

$L_1\&L_2$ are the two lines which passes through the point $(1,-1)$ and tangent to the curve $y=x^2-3x+2.$ Then which of the following point(s) lie on the hyperbola having $L_1\&L_2$ as its asymptotes and passing through the origin?

• A. $(2,-2)$
• B. $(\frac{-2}{3},-2)$
• C. $(\frac{8}{3},0)$
• D. $(-2,-2)$

Answer 1

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

The correct options are
A $(2,-2)$
B $(\frac{-2}{3},-2)$
C $(\frac{8}{3},0)$

The equation of line from $(1,-1)$ having slope $m$ is $y=mx-m-1$
This line is tangent to the curve $y=x^2-3x+2.$
$mx-m-1=x^2-3x+2x^2-(m+3)x+m+3=0△=0(m+3{)}^2-4(m+3)=0$
$\Rightarrow m=-3$ or $m=1$
The equation of tangents $\left(L_1\&L_2\right)$ are $3x+y-2=0\&x-y-2=0$
The equation of hyperbola and combined equation of its asymptotes differs only by constant.
The equation of hyperbola $(3x+y-2)(x-y-2)+\lambda =0$
It passes through $(0,0)\Rightarrow \lambda =-4$
$(3x+y-2)(x-y-2)-4=0$