Question Statement

Find the coordinates of the vertices and co-vertices of the following ellipses by differentiating the equation on both sides and solving for horizontal and vertical tangent lines:

(i) $2x^2+3y^2+2x-3y-5=0$ (ii) $\frac{(x-1{)}^2}{9}+\frac{(y+2{)}^2}{16}=1$ (iii) $\frac{1}{2}{x}^2+\frac{3}{4}{y}^2+x-y-5=0$ (iv) $x^2+2y^2-6x-4=0$

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Background and Explanation

To find the vertices and co-vertices of an ellipse, we identify the points where the tangent lines are horizontal or vertical. We use implicit differentiation to find the derivative $\frac{dy}{dx}$, setting it to $0$ for horizontal tangents and finding where it is undefined (denominator equals $0$) for vertical tangents. The longer axis connects the vertices, while the shorter axis connects the co-vertices.

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Solution

Part (i)

Given equation: $2x^2+3y^2+2x-3y-5=0$

1. Differentiate the equation w.r.t. $x$: $4x+6y\frac{dy}{dx}+2-3\frac{dy}{dx}=0$ $(6y-3)\frac{dy}{dx}=-(4x+2)$ $\frac{dy}{dx}=-\frac{4x+2}{6y-3}$ This represents the slope of the tangent at any point on the ellipse.

2. For horizontal tangents ($\frac{dy}{dx}=0$): $4x+2=0\Rightarrow x=-\frac{1}{2}$ Substitute $x=-\frac{1}{2}$ into the original equation: $2{\left(-\frac{1}{2}\right)}^2+3y^2+2\left(-\frac{1}{2}\right)-3y-5=0$ $\frac{1}{2}+3y^2-1-3y-5=0$ $3y^2-3y-\frac{11}{2}=0\text{ (Raw data uses }3y^2-3y-\frac{1}{2}=0)$ Using the quadratic formula as per raw data: $y=\frac{-(-3)\pm \sqrt{(-3{)}^2-4(3)\left(-\frac{1}{2}\right)}}{2(3)}=\frac{3\pm \sqrt{9+66}}{6}=\frac{3\pm \sqrt{75}}{6}=\frac{3\pm 5\sqrt{3}}{6}$ Points: $A\left(-\frac{1}{2},\frac{3-5\sqrt{3}}{6}\right)$ and $B\left(-\frac{1}{2},\frac{3+5\sqrt{3}}{6}\right)$. Distance $|AB|=\frac{10\sqrt{3}}{6}=\frac{5\sqrt{3}}{3}$.

3. For vertical tangents ($\frac{dy}{dx}=\infty$): Denominator must be zero: $6y-3=0\Rightarrow y=\frac{1}{2}$. Substitute $y=\frac{1}{2}$ into the original equation: $2x^2+3{\left(\frac{1}{2}\right)}^2+2x-3\left(\frac{1}{2}\right)-5=0$ $2x^2+2x-\frac{23}{4}=0$ $x=\frac{-2\pm \sqrt{(2{)}^2-4(2)\left(-\frac{23}{4}\right)}}{2(2)}=\frac{-2\pm \sqrt{4+46}}{4}=\frac{-2\pm 5\sqrt{2}}{4}$ Points: $C\left(\frac{-2-5\sqrt{2}}{4},\frac{1}{2}\right)$ and $D\left(\frac{-2+5\sqrt{2}}{4},\frac{1}{2}\right)$. Distance $|CD|=\frac{10\sqrt{2}}{4}=\frac{5\sqrt{2}}{2}$.

Since $|CD|>|AB|$, Vertices are $C,D$ and Co-vertices are $A,B$.

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Part (ii)

Given equation: $\frac{(x-1{)}^2}{9}+\frac{(y+2{)}^2}{16}=1$

1. Differentiate w.r.t. $x$: $\frac{2(x-1)}{9}+\frac{2(y+2)}{16}\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{16(x-1)}{9(y+2)}$

2. For horizontal tangents ($\frac{dy}{dx}=0$): $x-1=0\Rightarrow x=1$. Substitute $x=1$ into the equation: $0+\frac{(y+2{)}^2}{16}=1\Rightarrow (y+2{)}^2=16\Rightarrow y+2=\pm 4$ $y=2,-6$. Points: $A(1,-6)$ and $B(1,2)$. Distance $|AB|=8$.

3. For vertical tangents ($\frac{dy}{dx}=\infty$): $y+2=0\Rightarrow y=-2$. Substitute $y=-2$ into the equation: $\frac{(x-1{)}^2}{9}+0=1\Rightarrow (x-1{)}^2=9\Rightarrow x-1=\pm 3$ $x=4,-2$. Points: $C(-2,-2)$ and $D(4,-2)$. Distance $|CD|=6$.

Since $|AB|>|CD|$, Vertices are $A(1,-6),B(1,2)$ and Co-vertices are $C(-2,-2),D(4,-2)$.

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Part (iii)

Given equation: $\frac{1}{2}{x}^2+\frac{3}{4}{y}^2+x-y-5=0$

1. Differentiate w.r.t. $x$: $x+\frac{3}{2}y\frac{dy}{dx}+1-\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{x+1}{\frac{3}{2}y-1}$

2. For horizontal tangents: $x+1=0\Rightarrow x=-1$. Substitute into equation and multiply by 4: $3y^2-4y-22=0$. $y=\frac{4\pm \sqrt{16+264}}{6}=\frac{2\pm \sqrt{70}}{3}$ Points: $A\left(-1,\frac{2-\sqrt{70}}{3}\right),B\left(-1,\frac{2+\sqrt{70}}{3}\right)$. Distance $|AB|=\frac{2\sqrt{70}}{3}$.

3. For vertical tangents: $\frac{3}{2}y-1=0\Rightarrow y=\frac{2}{3}$. Substitute into equation and multiply by 6: $3x^2+6x-32=0$. $x=\frac{-6\pm \sqrt{36+384}}{6}=\frac{-3\pm \sqrt{105}}{3}$ Points: $C\left(\frac{-3-\sqrt{105}}{3},\frac{2}{3}\right),D\left(\frac{-3+\sqrt{105}}{3},\frac{2}{3}\right)$. Distance $|CD|=\frac{2\sqrt{105}}{3}$.

Since $|CD|>|AB|$, Vertices are $C,D$ and Co-vertices are $A,B$.

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Part (iv)

Given equation: $x^2+2y^2-6x-4=0$

1. Differentiate w.r.t. $x$: $2x+4y\frac{dy}{dx}-6=0\Rightarrow \frac{dy}{dx}=\frac{3-x}{2y}$

2. For horizontal tangents: $3-x=0\Rightarrow x=3$. Substitute $x=3$: $9+2y^2-18-4=0\Rightarrow 2y^2=13\Rightarrow y=\pm \sqrt{\frac{13}{2}}$. Points: $A(3,-\sqrt{\frac{13}{2}}),B(3,\sqrt{\frac{13}{2}})$. Distance $|AB|=\sqrt{26}$.

3. For vertical tangents: $2y=0\Rightarrow y=0$. Substitute $y=0$: $x^2-6x-4=0$. $x=\frac{6\pm \sqrt{36+16}}{2}=3\pm \sqrt{13}$ Points: $C(3-\sqrt{13},0),D(3+\sqrt{13},0)$. Distance $|CD|=2\sqrt{13}=\sqrt{52}$.

Since $|CD|>|AB|$, Vertices are $C,D$ and Co-vertices are $A,B$.

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Key Formulas or Methods Used

• Implicit Differentiation: $\frac{d}{dx}[f(x,y)=0]$
• Horizontal Tangent Condition: $\frac{dy}{dx}=0$
• Vertical Tangent Condition: $\frac{dy}{dx}\to \infty$ (Denominator of derivative $=0$)
• Quadratic Formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
• Distance Formula: $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

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Summary of Steps

1. Differentiate the ellipse equation implicitly to find the expression for $\frac{dy}{dx}$.
2. Set the numerator of $\frac{dy}{dx}$ to zero to find the $x$-coordinates (or $y$-coordinates) of points with horizontal tangents.
3. Set the denominator of $\frac{dy}{dx}$ to zero to find the coordinates of points with vertical tangents.
4. Substitute these values back into the original equation to find the corresponding pairs of points.
5. Calculate the distance between the pairs of points; the pair with the larger distance are the vertices, and the pair with the smaller distance are the co-vertices.