Common Tangent

The given equations are:- ${ y }^{ 2 }=8x$ $1$

$xy=-1$ $2$

The derivative of the equation $1$ is

$\frac { dy }{ dx } =\frac { 4 }{ y }$

The derivative of the equation $2$ is

$\frac { dy }{ dx } =\frac { -y }{ x }$

Consider the common tangent touches the parabola at $\left( { x }_{ 1 },{ y }_{ 1 } \right)$ and the hyperbola at $\left( { x }_{ 2 },{ y }_{ 2 } \right)$ . Then, the slopes at this point of the curves should be equal hence,

$\frac { 4 }{ { y }_{ 1 } } =\frac { { -y }_{ 2 } }{ { x }_{ 2 } }$

Now, the equation of tangent at both the curves at the respective points of tangency is as follows:

__At parabola:- __ $(y-{ y }_{ 1 })=\frac { 4 }{ { y }_{ 1 } } (x-{ x }_{ 1 })$

__At hyperbola :-__ $(y-{ y }_{ 21 })=\frac { { -y }_{ 2 } }{ { x }_{ 2 } } (x-{ x }_{ 2 })$

On solving and comparing both the equations we get the follows

$\frac { { y }_{ 1 } }{ { x }_{ 2 } } =\frac { -4 }{ { y }_{ 2 } } =\frac { { y }_{ 1 }^{ 2 }-4{ x }_{ 1 } }{ 2{ y }_{ 2 }{ x }_{ 2 } }$

Consider the $\frac { { y }_{ 1 } }{ { x }_{ 2 } } =\frac { { y }_{ 1 }^{ 2 }-4{ x }_{ 1 } }{ 2{ y }_{ 2 }{ x }_{ 2 } }$ part of the equality

$\frac { { y }_{ 1 } }{ { x }_{ 2 } } =\frac { { y }_{ 1 }^{ 2 }-4{ x }_{ 1 } }{ 2{ y }_{ 2 }{ x }_{ 2 } } \\ \Rightarrow \quad 2{ y }_{ 1 }{ y }_{ 2 }={ y }_{ 1 }^{ 2 }-4{ x }_{ 1 }\\ \Rightarrow \quad 2{ y }_{ 1 }{ y }_{ 2 }={ y }_{ 1 }^{ 2 }-\frac { { y }_{ 1 }^{ 2 } }{ 2 } (\because { y }_{ 1 }^{ 2 }=8{ x }_{ 1 })\\ \Rightarrow \quad 4{ y }_{ 1 }{ y }_{ 2 }={ y }_{ 1 }^{ 2 }\\ \Rightarrow \quad 4{ y }_{ 2 }={ y }_{ 1 }$

Now, consider

$\frac { { y }_{ 1 } }{ { x }_{ 2 } } =\frac { -4 }{ { y }_{ 2 } } \\ \Rightarrow \quad { y }_{ 1 }{ y }_{ 2 }=-4{ x }_{ 2 }\\ \Rightarrow \quad { x }_{ 2 }=-{ y }_{ 2 }^{ 2 }$

Using the above result in equation $2$ , we get ${ x }_{ 2 }{ y }_{ 2 }=-1\quad \Rightarrow \quad -{ y }_{ 2 }^{ 3 }=-1\quad \Rightarrow \quad { y }_{ 2 }=1$ Hence we get the following values:- ${ x }_{ 1 }=2{ ,\quad y }_{ 1 }=4.\quad { x }_{ 2 }=-1,\quad { y }_{ 2 }=1$

Using these we get the equation of tangent as $y=x+2\quad \Rightarrow \quad y-x-2=0$

$\therefore \quad a=1,\quad b=1,\quad c=-2$

So, $a-b+c=1+1-2=0$