Mathematics - Quadratic Equation Question with Solution | TestHub

Let $(\alpha ,\beta )$ be point on the curve such that the tangent drawn at $(\alpha ,\beta )$ passes through $(0$, 7)

$y^1=4x-4\Rightarrow y_{(\alpha ,\beta )}^1=4\alpha -4$

Tangent at $(\alpha ,\beta )$ is $y-\beta =(4\alpha -4)(x-\alpha )$ pass through $(0,-7)$

$\Rightarrow -7-\beta =(4\alpha -4)(0-\alpha )$

But $\beta =2{\alpha }^2-4\alpha -5\therefore$ It follows that ${\alpha }^2=1$

$\Rightarrow \alpha =\pm 1$

So, $x_1=1,x_2=-1$

So, $3x_1-2x_2=5$.