# Determine the points on the curve 5x² + 6xy +5y² =8 where the tangent is parallel to the line x-y= 1?

Sep 27, 2015

$\left(x , y\right) = \left(- \sqrt{2} , \sqrt{2}\right) , \left(\sqrt{2} , - \sqrt{2}\right)$

#### Explanation:

$x - y = 1 \implies$ write in slope-intercept form:
$y = x - 1 \implies$ slope = 1
$5 {x}^{2} + 6 x y + 5 {y}^{2} = 8 \implies$ implicit differentiation:
$10 x + 6 y + 6 x y ' + 10 y y ' = 0$
$y ' \left(3 x + 5 y\right) = - \left(5 x + 3 y\right)$
$y ' = - \frac{5 x + 3 y}{3 x + 5 y} \implies$ the slope of the curve, equate to 1:
$- \frac{5 x + 3 y}{3 x + 5 y} = 1 \implies$ solve for $y$ in terms of $x$:
$3 x + 5 y = - 5 x - 3 y$
$8 x + 8 y = 0$
$x + y = 0$
$y = - x \implies$ substitute in the equation of the curve, solve for $x$:
$5 {x}^{2} + 6 x \cdot \left(- x\right) + 5 \cdot {\left(- x\right)}^{2} = 8$
$5 {x}^{2} - 6 {x}^{2} + 5 {x}^{2} = 8$
$4 {x}^{2} = 8$
${x}^{2} = 2$
$x = - \sqrt{2} , \sqrt{2} \implies$ solve for $y :$
$y = \sqrt{2} , - \sqrt{2}$
hence the required points are:
$\left(x , y\right) = \left(- \sqrt{2} , \sqrt{2}\right) , \left(\sqrt{2} , - \sqrt{2}\right)$