Q. The point of contact of the line $ y=x-1 $ with $ 3x-4{{y}^{2}}=12 $ is
Jharkhand CECEJharkhand CECE 2014
Solution:
The equations of the line and hyperbola are
$ y=x-1 $ ... (i) $ 3{{x}^{2}}-4{{y}^{2}}=12 $ ... (ii)
From Eqs. (i) and (ii), we get
$ 3{{x}^{2}}-4{{(x-1)}^{2}}=12 $
$ \Rightarrow $ $ 3{{x}^{2}}-4({{x}^{2}}-2x+1)=12 $
or $ {{x}^{2}}-8x+16=0\Rightarrow x=4 $
From Eq. (i), $ y=3 $
Hence, required point of contact is $ (4,\,\,3) $ .
