Question Statement

Find the vertices of the following hyperbolas by differentiating the given equation and solving for horizontal or vertical tangent lines.

(i) $\frac{{\left(x-\frac{1}{2}\right)}^2}{3}-\frac{(y+3{)}^2}{5}=1$

(ii) $\frac{(y-1{)}^2}{2^2}-\frac{(x+1{)}^2}{19}=1$

(iii) $5x^2-5y^2+25x-5y+20=0$

(iv) $-x^2+y^2-10x-4y-28=0$

---

Background and Explanation

The vertices of a hyperbola are the points where the curve is at its "sharpest" turn. At these specific points, the tangent line is either perfectly horizontal (for vertical hyperbolas) or perfectly vertical (for horizontal hyperbolas). By using implicit differentiation to find the derivative $\frac{dy}{dx}$, we can identify these points by setting the slope to $0$ or finding where it is undefined ($\infty$).

---

Solution

Part (i)

The equation $\frac{{\left(x-\frac{1}{2}\right)}^2}{3}-\frac{(y+3{)}^2}{5}=1$ represents a horizontal hyperbola. In a horizontal hyperbola, the vertices occur where the tangent lines are vertical.

1. Differentiate the equation with respect to $x$: $\frac{1}{3}\left[2\left(x-\frac{1}{2}\right)\right]-\frac{1}{5}[2(y+3)]\frac{dy}{dx}=0$

2. Isolate the derivative $\frac{dy}{dx}$: $\frac{2}{5}(y+3)\frac{dy}{dx}=\frac{2}{3}\left(x-\frac{1}{2}\right)$ $\frac{dy}{dx}=\frac{\frac{2}{3}\left(x-\frac{1}{2}\right)}{\frac{2}{5}(y+3)}=\frac{5(x-\frac{1}{2})}{3(y+3)}$

3. Find the condition for a vertical tangent: A vertical tangent occurs when the slope is undefined ($\infty$), which happens when the denominator is zero: $3(y+3)=0⟹y+3=0⟹y=-3$

4. Substitute $y=-3$ back into the original equation to find $x$: $\frac{{\left(x-\frac{1}{2}\right)}^2}{3}-\frac{(-3+3{)}^2}{5}=1$ $\frac{{\left(x-\frac{1}{2}\right)}^2}{3}-0=1$ ${\left(x-\frac{1}{2}\right)}^2=3$ $x^2-x+\frac{1}{4}-3=0⟹x^2-x-\frac{11}{4}=0$

Using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$: $x=\frac{-(-1)\pm \sqrt{1-4(1)(-\frac{11}{4})}}{2}=\frac{1\pm \sqrt{1+11}}{2}=\frac{1\pm \sqrt{12}}{2}$ $x=\frac{1\pm 2\sqrt{3}}{2}$

Vertices: $\left(\frac{1-2\sqrt{3}}{2},-3\right)$ and $\left(\frac{1+2\sqrt{3}}{2},-3\right)$

---

Part (ii)

The equation $\frac{(y-1{)}^2}{4}-\frac{(x+1{)}^2}{19}=1$ represents a vertical hyperbola. The vertices occur where the tangent lines are horizontal.

1. Differentiate the equation with respect to $x$: $\frac{1}{4}[2(y-1)]\frac{dy}{dx}-\frac{1}{19}[2(x+1)]=0$

2. Isolate the derivative $\frac{dy}{dx}$: $\left(\frac{y-1}{2}\right)\frac{dy}{dx}=\frac{2}{19}(x+1)$ $\frac{dy}{dx}=\frac{4(x+1)}{19(y-1)}$

3. Find the condition for a horizontal tangent: A horizontal tangent occurs when $\frac{dy}{dx}=0$: $4(x+1)=0⟹x+1=0⟹x=-1$

4. Substitute $x=-1$ back into the original equation to find $y$: $\frac{(y-1{)}^2}{4}-\frac{(-1+1{)}^2}{19}=1$ $\frac{(y-1{)}^2}{4}-0=1⟹(y-1{)}^2=4$ $y^2-2y+1-4=0⟹y^2-2y-3=0$ $(y-3)(y+1)=0$ $y=3\text{ or }y=-1$

Vertices: $(-1,3)$ and $(-1,-1)$

---

Part (iii)

Given: $5x^2-5y^2+25x-5y+20=0$

1. Simplify the equation: Divide by 5: $x^2-y^2+5x-y+4=0$ Since the coefficient of $x^2$ is positive, this is a horizontal hyperbola; vertices have vertical tangents.

2. Differentiate with respect to $x$: $2x-2y\frac{dy}{dx}+5-\frac{dy}{dx}=0$ $(2y+1)\frac{dy}{dx}=2x+5$ $\frac{dy}{dx}=\frac{2x+5}{2y+1}$

3. Find the condition for a vertical tangent: Denominator must be zero: $2y+1=0⟹y=-\frac{1}{2}$

4. Substitute $y=-1/2$ back into the simplified equation: $x^2-(-\frac{1}{2}{)}^2+5x-(-\frac{1}{2})+4=0$ $x^2-\frac{1}{4}+5x+\frac{1}{2}+4=0⟹x^2+5x+\frac{17}{4}=0$

Using the quadratic formula: $x=\frac{-5\pm \sqrt{25-4(1)(\frac{17}{4})}}{2}=\frac{-5\pm \sqrt{25-17}}{2}=\frac{-5\pm \sqrt{8}}{2}$ $x=\frac{-5\pm 2\sqrt{2}}{2}$

Vertices: $\left(\frac{-5-2\sqrt{2}}{2},-\frac{1}{2}\right)$ and $\left(\frac{-5+2\sqrt{2}}{2},-\frac{1}{2}\right)$

---

Part (iv)

Given: $-x^2+y^2-10x-4y-28=0$

1. Identify orientation: Since the coefficient of $y^2$ is positive, this is a vertical hyperbola; vertices have horizontal tangents.

2. Differentiate with respect to $x$: $-2x+2y\frac{dy}{dx}-10-4\frac{dy}{dx}=0$ $(2y-4)\frac{dy}{dx}=2x+10$ $\frac{dy}{dx}=\frac{2x+10}{2y-4}=\frac{x+5}{y-2}$

3. Find the condition for a horizontal tangent: Numerator must be zero: $x+5=0⟹x=-5$

4. Substitute $x=-5$ back into the original equation: $-(-5{)}^2+y^2-10(-5)-4y-28=0$ $-25+y^2+50-4y-28=0⟹y^2-4y-3=0$

Using the quadratic formula: $y=\frac{-(-4)\pm \sqrt{(-4{)}^2-4(1)(-3)}}{2}=\frac{4\pm \sqrt{16+12}}{2}=\frac{4\pm \sqrt{28}}{2}$ $y=\frac{4\pm 2\sqrt{7}}{2}=2\pm \sqrt{7}$

Vertices: $(-5,2-\sqrt{7})$ and $(-5,2+\sqrt{7})$

---

Key Formulas or Methods Used

• Implicit Differentiation: Differentiating both sides of an equation with respect to $x$, treating $y$ as a function of $x$.
• Horizontal Tangent Condition: $\frac{dy}{dx}=0$.
• Vertical Tangent Condition: $\frac{dy}{dx}$ is undefined (denominator equals zero).
• Quadratic Formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.
• Hyperbola Orientation:
  • If $x^2$ term is positive: Horizontal (Vertical tangents at vertices).
  • If $y^2$ term is positive: Vertical (Horizontal tangents at vertices).

---

Summary of Steps

1. Determine if the hyperbola is horizontal or vertical based on the signs of the $x^2$ and $y^2$ coefficients.
2. Differentiate the equation implicitly to find the expression for $\frac{dy}{dx}$.
3. Set the derivative to $0$ (for vertical hyperbolas) or set the denominator to $0$ (for horizontal hyperbolas).
4. Solve for the resulting variable ($x$ or $y$).
5. Plug that value back into the original equation to find the coordinates of the vertices.