Question

Find the points common to the hyperbola $25x^2-9y^2=225$ and the straight line $25x+12y-45=0$.

Answer 1

Rewriting the equation of straight line, we get $x=\frac{45-12y}{25}$

Putting the value of $x$ in equation of hyperbola and solving for $y,$ we get

$9y^2+120y+400=0$

$\Rightarrow (3y+20{)}^2=0$

$\Rightarrow y=-\frac{20}{3}$

Solving for $x,$ we get
$x=5$

The line is tangent to the hyperbola and $(5,-\frac{20}{3})$ satisfies both the equations, hence $(5,-\frac{20}{3})$ is common to both.