Chapter 4, Problem 72RE

a)

To determine

To find: The coordinates of points $A$ and $B$ of given ellipse where tangent lines are horizontal.

Answer to Problem 72RE

Points are $(\sqrt{2},2\sqrt{2}),(-\sqrt{2},-2\sqrt{2})$

Explanation of Solution

Given: Equation of ellipse $2x^2-2xy+y^2-4=0$

Concept Used: Differentiation with respect to $x$

Calculation: We have equation $2x^2-2xy+y^2-4=0$

Differentiate above equation with respect to $x$ we get,

  $4x-\left(2x\frac{dy}{dx}+2y\right)+2y\frac{dy}{dx}=0$

  $4x-2x\frac{dy}{dx}-2y+2y\frac{dy}{dx}=0$

  $\frac{dy}{dx}\left(2y-2x\right)=2y-4x$ $C$

  $\frac{dy}{dx}=\frac{2y-4x}{2y-2x}=\frac{y-2x}{y-x}$

Tangent will be horizontal when numerator of the derivative equals to 0 or $\frac{dy}{dx}=0$

  $y-2x=0$

  $y=2x$

Put $y=2x$ in original equation, we get

  $2x^2-2x\left(2x\right)+{\left(2x\right)}^2-4=0$

  $2x^2-4=0$

  $x=\pm \sqrt{2}$

Now put $x=\pm \sqrt{2}$ in $y=2x$ we get,

  $y=\pm 2\sqrt{2}$

Conclusion: Hence, the points where the tangent lines are horizontal are $(\sqrt{2},2\sqrt{2}),(-\sqrt{2},-2\sqrt{2})$

b)

To determine

To find: The coordinates of points $C$ and $D$ of given ellipse where tangent lines are vertical.

Answer to Problem 72RE

Points are $\left(2,2\right),\left(-2,-2\right)$

Explanation of Solution

Given: Equation of ellipse $2x^2-2xy+y^2-4=0$

Concept Used: Differentiation with respect to $x$

Calculation: We have equation $2x^2-2xy+y^2-4=0$

Differentiate above equation with respect to $x$

  $4x-\left(2x\frac{dy}{dx}+2y\right)+2y\frac{dy}{dx}=0$

  $4x-2x\frac{dy}{dx}-2y+2y\frac{dy}{dx}=0$

  $\frac{dy}{dx}\left(2y-2x\right)=2y-4x$ $C$

  $\frac{dy}{dx}=\frac{2y-4x}{2y-2x}=\frac{y-2x}{y-x}$

Tangent will be vertical when denominator of $\frac{dy}{dx}=0$

  $y-x=0$

  $y=x$

Now, substitute the $y=x$ in $2x^2-2xy+y^2-4=0$ and we get,

  $2x^2-2x(x)+{(x)}^2-4=0$

  $x^2-4=0$

  $x^2=4$

  $x=\pm 2$

Now put $x=\pm 2$ in $y=x$ we get,

  $x=2,y=2$

  $x=-2,y=-2$

Conclusion: Hence, the points where the tangent is vertical are $\left(2,2\right),\left(-2,-2\right)$