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 To find  the point on the graph where the
tangent line is || to y = x+7.

A line y = mx+c is a secant
line to the curve x^2+2y^2 = 1 intercepting at two points as the curve is a 2nd degree
equation.

But suppose this secant line is parallel to the
line y = x+7, then m = 1.

So we consider the possibilty of
the tangent line from the family of secant lines y = x+c  whose  slope is
1.

Substituting y = x+c in (1) we get: x^2+ 2(x+c)^2 =
1.

x^2+2x^2+4cx +2c^2 -1 =
0.

3x^2+4cx+2c^2-1 =
0.....(1).

When the discriminant of eq(1) . that is,
 (4c)^2 - 4(3)(2c^2-1) = 0 , the secant line makes only one single coinciding double
point intercept. So the secant line becomes tangent  for this value of
c.

Therefore (4c)^2 - 4*3*(2c^2-1) =
0.

16c^2  -24c^2 + 12 =
0.

12 = 8c^2.

c^2 = 
(3/2).

c^2 = 3/2.

 c =
sqrt(3/2), or

c =
-sqrt(3/2).

From (1)  x =  {-4c +or-
sqrt(discriminant)}/(2*3)

x = -2c/3  , as discriminant is
zero.

x1 = -(2/3)(sqrt(3/2) ,
or x2 = -(2/3) sqrt(-5/6) = (2/3)
sqrt(3/2).

When x = x1 = -(2/3) sqrt(3/2)
, y = y1 = (-2/3)(sqrt(3/2) +(-sqrt(3/2)) =
(-2/3-1)sqrt(3/2) =
-(5/3)sqrt(3/2).

When x = x2 =
(2/3)sqrt(3/2), y=y2 = x+sqrt(3/2) = (2/3)
sqrt(3/2)+sqrt(3/2) = (2/3+1) sqrt(3/2) = (5/3)
sqrt(3/2).

Therefore there are
2 points, {-(2/3)sqrt(3/2) , -(5/3)sqrt(3/2)} and {(2/3)sqrt(3/2) , (5/3)sqrt(3/2)} 
where the tangent lines are parallel to y =
x+7.

The two
tangent lines are  y = x+sqrt(3/2) or y = x-sqrt(3/2).