Question

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

Answer 1

`(x, y)=(-sqrt2, sqrt2) , (sqrt2, -sqrt2)`

Explanation:

`x-y=1=>` write in slope-intercept form:
`y=x-1=>` slope = 1
`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)`