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

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

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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neela | High School Teacher | (Level 3) Valedictorian

Posted on

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).

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