zgribestika

2021-12-28

Obtain the Differential Equation from the family of ellipses centered at (h,k) whose semi-major axis is twice the semi-minor axis.

xandir307dc

Beginner2021-12-29Added 35 answers

Equation from family of ellipses whose centered at (h, k) and semi mayor axis is twice the semi minor axis

mayor axis$=a$

minor axis$=b$

General equation of ellipse:

$\frac{{(x-h)}^{2}}{{a}^{2}}+\frac{{(y-k)}^{2}}{{b}^{2}}=1$ (1)

Given condition:

Semi mayor axis is twice the semi minor

$\frac{a}{2}=2\left(\frac{b}{2}\right)\Rightarrow \frac{a}{2}=b$ (put in 1)

Then 1 becomes$\frac{{(x-h)}^{2}}{{a}^{2}}+\frac{{(y-k)}^{2}}{{\left(\frac{a}{2}\right)}^{2}}=1$

$\Rightarrow \frac{{(x-h)}^{2}}{{a}^{2}}+\frac{4{(y-k)}^{2}}{{a}^{2}}=1$

$\Rightarrow {(x-h)}^{2}+4{(y-k)}^{2}={a}^{2}$

$2(x-h)+g(y-k)\frac{dy}{dx}=0$

$\frac{dy}{dx}=\frac{-2(x+h)}{8(y-k)}\Rightarrow \frac{-(x-h)}{4(y-k)}=\frac{dy}{dx}$

mayor axis

minor axis

General equation of ellipse:

Given condition:

Semi mayor axis is twice the semi minor

Then 1 becomes

servidopolisxv

Beginner2021-12-30Added 27 answers

Step 1

Given data is:

The semi-major axis is twice the semi-minor axis and the center of the ellipse is (h,k)

To find the differential equation from the family of the ellipse.

Step 2

Let a be the major axis and b is the minor-axis

The center of the ellipse is (h,k), therefore the equation of the ellipse is:

$\frac{{(x-h)}^{2}}{{a}^{2}}+\frac{{(y-k)}^{2}}{{b}^{2}}=1$ (1)

From the question,

The semi-major axis is twice the semi-minor axis, therefore

$a=2b$

From equation (1),

$\frac{{(x-h)}^{2}}{{\left(2b\right)}^{2}}+\frac{{(y-k)}^{2}}{{b}^{2}}=1$

$\frac{{(x-h)}^{2}}{4{b}^{2}}+\frac{{(y-k)}^{2}}{{b}^{2}}=1$

$(x-h)}^{2}+4{(y-k)}^{2}=4{b}^{2$ (2)

Step 3

Differentiate equation (2) with respect to x,

$2(x-h)+4\times 2(y-k){y}^{\prime}=0$

$2(x-h)+8(y-k){y}^{\prime}=0$

Again differentiate with respect to x:

$2\times 1+8\{(y-k)y{y}^{\prime}\times {y}^{\prime}\}=0$

$8(y-k)y8{\left({y}^{\prime}\right)}^{2}=-1$

$(y-k)y{\left({y}^{\prime}\right)}^{2}=-\frac{1}{8}$

Thus, the differential equation is$(y-k)y{\left({y}^{\prime}\right)}^{2}=-\frac{1}{8}$

Given data is:

The semi-major axis is twice the semi-minor axis and the center of the ellipse is (h,k)

To find the differential equation from the family of the ellipse.

Step 2

Let a be the major axis and b is the minor-axis

The center of the ellipse is (h,k), therefore the equation of the ellipse is:

From the question,

The semi-major axis is twice the semi-minor axis, therefore

From equation (1),

Step 3

Differentiate equation (2) with respect to x,

Again differentiate with respect to x:

Thus, the differential equation is

user_27qwe

Skilled2022-01-05Added 375 answers

\(\begin{array}{}\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1 (1) \a=2b \\frac{(x-h)^{2}}{(2b)^{2}}+\frac{(y-k)^{2}}{b^{2}}=1 \\frac{(x-h)^{2}}{4b^{2}}+\frac{(y-k)^{2}}{b^{2}}=1 \(x-h)^{2}+4(y-k)^{2}=4b^{2} (2) \2(x-h)+4 \times 2(y-k)y=0

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