Kelton Bailey

2022-10-01

Finding the foci and vertices of an ellipse.

How would you find the foci and vertices of the following ellipse: $\frac{2{x}^{2}}{15}+\frac{8{y}^{2}}{45}-\frac{2\sqrt{3}}{45}xy=1?$

How would you find the foci and vertices of the following ellipse: $\frac{2{x}^{2}}{15}+\frac{8{y}^{2}}{45}-\frac{2\sqrt{3}}{45}xy=1?$

Matteo Estes

Beginner2022-10-02Added 9 answers

Step 1

Notice first of all that your ellipse is centered at $O=(0,0)$ (because if (x,y) belongs to the ellipse, then also (-x, -y) belongs to it).

To find the axes of the ellipse, notice that if $P=(x,y)$ is a vertex, then the tangent at P is perpendicular to PO, that is ${y}^{\prime}(y/x)=-1$. You can compute y′ by differentiating the equation of the ellipse:

$\frac{4}{15}x+\frac{16}{45}y{y}^{\prime}-\frac{2\sqrt{3}}{45}y-\frac{2\sqrt{3}}{45}x{y}^{\prime}=0,$

whence: ${y}^{\prime}=\frac{2\sqrt{3}y-12x}{16y-2\sqrt{3}x}.$

Step 2

The above condition ${y}^{\prime}(y/x)=-1$ implies then that the coordinates of a vertex are related by:

$\frac{y}{x}=\frac{-1\pm 2}{\sqrt{3}}.$

Plugging that into the ellipse equation you can get the coordinates of the vertices, and then of course those of the foci.

Notice first of all that your ellipse is centered at $O=(0,0)$ (because if (x,y) belongs to the ellipse, then also (-x, -y) belongs to it).

To find the axes of the ellipse, notice that if $P=(x,y)$ is a vertex, then the tangent at P is perpendicular to PO, that is ${y}^{\prime}(y/x)=-1$. You can compute y′ by differentiating the equation of the ellipse:

$\frac{4}{15}x+\frac{16}{45}y{y}^{\prime}-\frac{2\sqrt{3}}{45}y-\frac{2\sqrt{3}}{45}x{y}^{\prime}=0,$

whence: ${y}^{\prime}=\frac{2\sqrt{3}y-12x}{16y-2\sqrt{3}x}.$

Step 2

The above condition ${y}^{\prime}(y/x)=-1$ implies then that the coordinates of a vertex are related by:

$\frac{y}{x}=\frac{-1\pm 2}{\sqrt{3}}.$

Plugging that into the ellipse equation you can get the coordinates of the vertices, and then of course those of the foci.

seguitzla

Beginner2022-10-03Added 4 answers

Step 1

Consider a rotation: $x=X\mathrm{cos}\alpha -Y\mathrm{sin}\alpha ,\phantom{\rule{2em}{0ex}}y=X\mathrm{sin}\alpha +Y\mathrm{cos}\alpha $

Then your ellipse becomes $6(X\mathrm{cos}\alpha -Y\mathrm{sin}\alpha {)}^{2}+8(X\mathrm{sin}\alpha +Y\mathrm{cos}\alpha {)}^{2}-2\sqrt{3}(X\mathrm{cos}\alpha -Y\mathrm{sin}\alpha )(X\mathrm{sin}\alpha +Y\mathrm{cos}\alpha )=45$

The term in XY has coefficient $4\mathrm{cos}\alpha \mathrm{sin}\alpha -2\sqrt{3}{\mathrm{cos}}^{2}\alpha +2\sqrt{3}{\mathrm{sin}}^{2}\alpha $ which you want to be vanishing; dividing by ${\mathrm{cos}}^{2}\alpha $ we get $2\sqrt{3}{\mathrm{tan}}^{2}\alpha +4\mathrm{tan}\alpha -2\sqrt{3}=0$ so $\mathrm{tan}\alpha =\frac{-2+4}{2\sqrt{3}}=\frac{1}{\sqrt{3}}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\mathrm{tan}\alpha =\frac{-2-4}{2\sqrt{3}}=-\sqrt{3}$

Step 2

So we can take $\alpha =\pi /6$ and the equation of the ellipse becomes

${X}^{2}(6{\mathrm{cos}}^{2}\alpha +8{\mathrm{sin}}^{2}\alpha -2\sqrt{3}\mathrm{cos}\alpha \mathrm{sin}\alpha )+{Y}^{2}(6{\mathrm{sin}}^{2}\alpha +8{\mathrm{cos}}^{2}\alpha +2\sqrt{3}\mathrm{cos}\alpha \mathrm{sin}\alpha )=45$

that is $5{X}^{2}+9{Y}^{2}=45$ or $\frac{{X}^{2}}{9}+\frac{{Y}^{2}}{5}=1$.

Find foci and vertices, then use the inverse rotation.

Consider a rotation: $x=X\mathrm{cos}\alpha -Y\mathrm{sin}\alpha ,\phantom{\rule{2em}{0ex}}y=X\mathrm{sin}\alpha +Y\mathrm{cos}\alpha $

Then your ellipse becomes $6(X\mathrm{cos}\alpha -Y\mathrm{sin}\alpha {)}^{2}+8(X\mathrm{sin}\alpha +Y\mathrm{cos}\alpha {)}^{2}-2\sqrt{3}(X\mathrm{cos}\alpha -Y\mathrm{sin}\alpha )(X\mathrm{sin}\alpha +Y\mathrm{cos}\alpha )=45$

The term in XY has coefficient $4\mathrm{cos}\alpha \mathrm{sin}\alpha -2\sqrt{3}{\mathrm{cos}}^{2}\alpha +2\sqrt{3}{\mathrm{sin}}^{2}\alpha $ which you want to be vanishing; dividing by ${\mathrm{cos}}^{2}\alpha $ we get $2\sqrt{3}{\mathrm{tan}}^{2}\alpha +4\mathrm{tan}\alpha -2\sqrt{3}=0$ so $\mathrm{tan}\alpha =\frac{-2+4}{2\sqrt{3}}=\frac{1}{\sqrt{3}}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\mathrm{tan}\alpha =\frac{-2-4}{2\sqrt{3}}=-\sqrt{3}$

Step 2

So we can take $\alpha =\pi /6$ and the equation of the ellipse becomes

${X}^{2}(6{\mathrm{cos}}^{2}\alpha +8{\mathrm{sin}}^{2}\alpha -2\sqrt{3}\mathrm{cos}\alpha \mathrm{sin}\alpha )+{Y}^{2}(6{\mathrm{sin}}^{2}\alpha +8{\mathrm{cos}}^{2}\alpha +2\sqrt{3}\mathrm{cos}\alpha \mathrm{sin}\alpha )=45$

that is $5{X}^{2}+9{Y}^{2}=45$ or $\frac{{X}^{2}}{9}+\frac{{Y}^{2}}{5}=1$.

Find foci and vertices, then use the inverse rotation.

The distance between the centers of two circles C1 and C2 is equal to 10 cm. The circles have equal radii of 10 cm.

A part of circumference of a circle is called

A. Radius

B. Segment

C. Arc

D. SectorThe perimeter of a basketball court is 108 meters and the length is 6 meters longer than twice the width. What are the length and width?

What are the coordinates of the center and the length of the radius of the circle represented by the equation ${x}^{2}+{y}^{2}-4x+8y+11=0$?

Which of the following pairs of angles are supplementary?

128,62

113,47

154,36

108,72What is the surface area to volume ratio of a sphere?

An angle which measures 89 degrees is a/an _____.

right angle

acute angle

obtuse angle

straight angleHerman drew a 4 sided figure which had only one pair of parallel sides. What could this figure be?

Trapezium

Parallelogram

Square

RectangleWhich quadrilateral has: All sides equal, and opposite angles equal?

Trapezium

Rhombus

Kite

RectangleKaren says every equilateral triangle is acute. Is this true?

Find the number of lines of symmetry of a circle.

A. 0

B. 4

C. 2

D. InfiniteThe endpoints of a diameter of a circle are located at (5,9) and (11, 17). What is the equation of the circle?

What is the number of lines of symmetry in a scalene triangle?

A. 0

B. 1

C. 2

D. 3How many diagonals does a rectangle has?

A quadrilateral whose diagonals are unequal, perpendicular and bisect each other is called a.

A. rhombus

B. trapezium

C. parallelogram