Chapter 4, Problem 71RE

a)

To determine

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

Answer to Problem 71RE

The coordinate points are $\left(1,-1\right),\left(-1,1\right)$ .

Explanation of Solution

Given: Equation of hyperbola $4x^2+8xy+y^2+3=0$ .

Concept Used: Differentiation with respect to $x$

Calculation: We have equation $4x^2+8xy+y^2+3=0$ .

Differentiate above equation with respect to $x$ :

  $\frac{d\left(4x^2\right)}{dx}+\frac{d\left(8x\right)}{dx}+8x.\frac{dy}{dx}+\frac{d\left(y^2\right)}{dy}.\frac{dy}{dx}+0=0$

  $8x+8y+8x.\frac{dy}{dx}+2y.\frac{dy}{dx}=0$

  $8x.\frac{dy}{dx}+2y.\frac{dy}{dx}=-\left(8x+8y\right)$

  $\left(8x+2y\right).\frac{dy}{dx}=-\left(8x+8y\right)$

Tangent will be horizontal when $\frac{dy}{dx}=0$

  $\Rightarrow 0=-8(x+y)$

  $x+y=0$

  $y=-x$

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

  $4x^2+8x(-x)+(-x^2)+3=0$

  $4x^2-8x+x^2+3=0$

  $-3x^2+3=0$

  $x^2=\pm 1$

Conclusion: Hence, the points where the tangent lines are horizontal are $\left(1,-1\right),\left(-1,1\right)$

b)

To determine

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

Answer to Problem 71RE

The coordinate points are $\left(-\frac{1}{2},2\right),\left(\frac{1}{2},-2\right)$ .

Explanation of Solution

Given: Equation of hyperbola $4x^2+8xy+y^2+3=0$ .

Concept Used: Differentiation with respect to $x$

Calculation: The given equation $4x^2+8xy+y^2+3=0$ .

Differentiate above equation with respect to $x$

  $\frac{d\left(4x^2\right)}{dx}+\frac{d\left(8x\right)}{dx}+8x.\frac{dy}{dx}+\frac{d\left(y^2\right)}{dy}.\frac{dy}{dx}+0=0$

  $8x+8y+8x.\frac{dy}{dx}+2y.\frac{dy}{dx}=0$

  $8x.\frac{dy}{dx}+2y.\frac{dy}{dx}=-\left(8x+8y\right)$

  $\left(8x+2y\right).\frac{dy}{dx}=-\left(8x+8y\right)$

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

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

  $\begin{array}{l}\Rightarrow 4x+y=0\\ y=-4x\end{array}$

Now, substitute the $y=-4x$ in $4x^2+8xy+y^2+3=0$ and we get,

  $4x^2+8xy+y^2+3=0$

  $4x^2+8x(-4x)+(-4x^2)+3=0$

  $4x^2-32x^2+16x^2+3=0$

  $x^2=\frac{1}{4}$

  $x=\pm \frac{1}{2}$

Now put $x=\pm \frac{1}{2}$ in $y=-4x$ we get,

  $x=\frac{1}{2},y=-2$

  $x=-\frac{1}{2},y=2$

Conclusion: Hence, the points where the tangent is vertical are $\left(-\frac{1}{2},2\right),\left(\frac{1}{2},-2\right)$