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

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:
$\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1$ (1)
Given condition:
Semi mayor axis is twice the semi minor
$\frac{a}{2}=2\left(\frac{b}{2}\right)⇒\frac{a}{2}=b$ (put in 1)
Then 1 becomes $\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{\left(\frac{a}{2}\right)}^{2}}=1$
$⇒\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{4{\left(y-k\right)}^{2}}{{a}^{2}}=1$
$⇒{\left(x-h\right)}^{2}+4{\left(y-k\right)}^{2}={a}^{2}$
$2\left(x-h\right)+g\left(y-k\right)\frac{dy}{dx}=0$
$\frac{dy}{dx}=\frac{-2\left(x+h\right)}{8\left(y-k\right)}⇒\frac{-\left(x-h\right)}{4\left(y-k\right)}=\frac{dy}{dx}$

servidopolisxv

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:
$\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1$ (1)
From the question,
The semi-major axis is twice the semi-minor axis, therefore
$a=2b$
From equation (1),
$\frac{{\left(x-h\right)}^{2}}{{\left(2b\right)}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1$
$\frac{{\left(x-h\right)}^{2}}{4{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1$
${\left(x-h\right)}^{2}+4{\left(y-k\right)}^{2}=4{b}^{2}$ (2)
Step 3
Differentiate equation (2) with respect to x,
$2\left(x-h\right)+4×2\left(y-k\right){y}^{\prime }=0$
$2\left(x-h\right)+8\left(y-k\right){y}^{\prime }=0$
Again differentiate with respect to x:
$2×1+8\left\{\left(y-k\right)y{y}^{\prime }×{y}^{\prime }\right\}=0$
$8\left(y-k\right)y8{\left({y}^{\prime }\right)}^{2}=-1$
$\left(y-k\right)y{\left({y}^{\prime }\right)}^{2}=-\frac{1}{8}$
Thus, the differential equation is $\left(y-k\right)y{\left({y}^{\prime }\right)}^{2}=-\frac{1}{8}$

user_27qwe