# At what points on the graph of x^2+2y^2=1 is the tangent line parallel to the line y=x+7? x^2 + 2y^2 = 1

Let us determine the tangent line equation.

Differentiate with respect to x:

==> 2x + 4yy' = 0

==> y'= -2x/4y = -x/2y

But we know that slope of the tangent line is parallel to the line y= x+ 7.

Then the slope for the...

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x^2 + 2y^2 = 1

Let us determine the tangent line equation.

Differentiate with respect to x:

==> 2x + 4yy' = 0

==> y'= -2x/4y = -x/2y

But we know that slope of the tangent line is parallel to the line y= x+ 7.

Then the slope for the tangent line = the slope for the line y= x+ 7 which is 1:

==> y' = 1

==>y' = -x/2y = 1

==> -x = 2y

==> x= -2y

Then the point is when x= -2y.

Subsitute in the cricle equation to determine te point where x= -2y:

==> 4y^2 + 2y^2 = 1

==> 6y^2 = 1

==> y^2 = 1/6

==> y= +- 1/sqrt6

==> x= +-2/sqrt6

Then there are 2 tangent lines parallel to the line y= x+ 7  at the points    ( 2/sqrt6, -1/sqrt6)    ( -2/qrt6, 1/sqrt6)

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