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

Name: Obtain the Differential Equation from the family of ellipses centered at (h,k) whose semi-major axis is twice the semi-minor axis.
Uploaded: 2022-02-23T17:22:57+00:00
Description: Obtain the Differential Equation from the family of ellipses centered at (h,k) whose semi-major axis is twice the semi-minor axis.

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xandir307dc · Accepted Answer

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:  (x−h)2a2+(y−k)2b2=1 (1)  Given condition:  Semi mayor axis is twice the semi minor   a2=2(b2)⇒a2=b (put in 1)  Then 1 becomes (x−h)2a2+(y−k)2(a2)2=1  ⇒(x−h)2a2+4(y−k)2a2=1  ⇒(x−h)2+4(y−k)2=a2  2(x−h)+g(y−k)dydx=0  dydx=−2(x+h)8(y−k)⇒−(x−h)4(y−k)=dydx

servidopolisxv · Accepted Answer

Step 1Given 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 2Let a be the major axis and b is the minor-axisThe center of the ellipse is (h,k), therefore the equation of the ellipse is:(x−h)2a2+(y−k)2b2=1 (1)From the question,The semi-major axis is twice the semi-minor axis, thereforea=2bFrom equation (1),(x−h)2(2b)2+(y−k)2b2=1(x−h)24b2+(y−k)2b2=1(x−h)2+4(y−k)2=4b2 (2)Step 3Differentiate equation (2) with respect to x,2(x−h)+4×2(y−k)y′=02(x−h)+8(y−k)y′=0Again differentiate with respect to x:2×1+8{(y−k)yy′×y′}=08(y−k)y8(y′)2=−1(y−k)y(y′)2=−18Thus, the differential equation is (y−k)y(y′)2=−18

user_27qwe · Accepted Answer

(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

Obtain the Differential Equation from the family of ellipses centered

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