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

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

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Answer 1 · 2015-09-27T02:38:41

$(x, y) = (- \sqrt{2}, \sqrt{2}), (\sqrt{2}, - \sqrt{2})$ $(x, y)=(-sqrt2, sqrt2) , (sqrt2, -sqrt2)$

Explanation:

$x - y = 1 \Rightarrow$ $x-y=1=>$ write in slope-intercept form:
$y = x - 1 \Rightarrow$ $y=x-1=>$ slope = 1
$5 x^{2} + 6 x y + 5 y^{2} = 8 \Rightarrow$ $5x^2+ 6xy +5y^2 =8=>$ implicit differentiation:
$10x+6y+6xy'+10yy'=0$
$y'(3x+5y)=-(5x+3y)$
$y'=-(5x+3y)/(3x+5y)=>$ the slope of the curve, equate to 1:
$-(5x+3y)/(3x+5y)=1=>$ solve for $y$ in terms of $x$ :
$3x+5y=-5x-3y$
$8x+8y=0$
$x+y=0$
$y=-x=>$ substitute in the equation of the curve, solve for $x$ :
$5x^2+6x*(-x)+5*(-x)^2=8$
$5x^2-6x^2+5x^2=8$
$4x^2=8$
$x^2=2$
$x=-sqrt2 , sqrt2=>$ solve for $y:$
$y=sqrt2 , -sqrt2$
hence the required points are:
$(x, y)=(-sqrt2, sqrt2) , (sqrt2, -sqrt2)$

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

1 Answer

Explanation:

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