#### To determine

**To find:**

Pointonthe ellipse x2+2y2=1 where the tangent line has slope 1.

#### Answer

23,-16 and (-23, 16)

#### Explanation

**1) Formula:**

i. Sum rule:

ddxfx+gx=ddxfx+ddxg(x)

ii. ddxsinx=cosx

iii. ddxcosx=-sinx

iv. Equation of tangent line to the curve y=f(x) at point (a,f(a)) is, (y-f(a))=f’(a)(x-a)

**2) Given:**

x2+2y2=1

**Slope=1**

**3) Calculation:**

Differentiate given equation of curve with respect to x,

ddx(x2+2y2)=ddx1

By using sum rule, constant function rule and power rule,

2x+4y.dydx=0

Solving for dy/dx

dydx=-x2y

The slope of tangent at a given point is value of derivative at that point.

But here slope is given as 1, so setting dy/dx equal to 1 we have.

-x2y=1

⇒x=-2y

Thus it follows that if tangent at (x, y) has slope 1 then x=-2y

Using this substitute for x in equation of ellipse, thus we have

(-2y)2+2y2=1

4y2+2y2=1

6y2=1

y2=16

⇒y=±16

Given a value of y we can find value of x by substituting this value in x=-2y

Therefore, points are

23,-16 and (-23, 16)

Notice that in the above we used

26=46=46=23

**Conclusion:**

The points where the given ellipse has tangent with slope 1 are

23,-16 and -23, 16