Question

Find the differential equation representing the family of ellipse having foci either on the x-axis or on the y-axis, centre at the origin and passing through points $\left( {0,3} \right)$
A. $xyy' + {y^2} - 9 = 0$
B. $x + yy'' = 0$
C. $xyy'' + x{\left( {y'} \right)^2} - yy' = 0$
D. $xyy' - {y^2} + 9 = 0$

Answer 1

Use the standard equation of an ellipse
$\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ where a>b . Then put the value of the points through which it passes and differentiate with respect to x. Adjust and simplify the equation to get the answer.

Complete step by step solution:
Here it is given that the ellipse has foci on either x-axis or y-axis and the centre is at origin then, we know that the standard equation of ellipse is given as
$\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ where a>b
Now since it is given that the ellipse passes through points $\left( {0,3} \right)$ then the equation of ellipse becomes-
$\Rightarrow \dfrac{0}{{{a^2}}} + \dfrac{{{{\left( 3 \right)}^3}}}{{{b^2}}} = 1$
On solving we get-
$\Rightarrow \dfrac{9}{{{b^2}}} = 1$
On further solving, we get-
$\Rightarrow {b^2} = 9 \Rightarrow b = \sqrt 9$
So we get the value of b-
$\Rightarrow b = \pm 3$
Now on putting the value of b in given eq. of ellipse, we get-
$\Rightarrow \dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{3^2}}} = 1$
On solving we get,
$\Rightarrow \dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{9} = 1$ --- (i)
From this equation we can also derive,
$\Rightarrow \dfrac{1}{{{a^2}}} = \dfrac{1}{{{x^2}}}\left( {1 - \dfrac{{{y^2}}}{9}} \right)$ -- (ii)
Now on differentiating eq. (i) w.r.t. x , we get
$\Rightarrow \dfrac{d}{{dx}}\left[ {\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{9} = 1} \right]$
$\Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{{x^2}}}{{{a^2}}}} \right) + \dfrac{d}{{dx}}\left( {\dfrac{{{y^2}}}{9}} \right) = \dfrac{d}{{dx}}\left( 1 \right)$
On simplifying we get,
$\Rightarrow \dfrac{1}{{{a^2}}}\dfrac{d}{{dx}}\left( {{x^2}} \right) + \dfrac{1}{9}\dfrac{d}{{dx}}\left( {{y^2}} \right) = \dfrac{d}{{dx}}\left( 1 \right)$
Now we know that differentiation of constant is zero so,
$\Rightarrow \dfrac{1}{{{a^2}}}\dfrac{d}{{dx}}\left( {{x^2}} \right) + \dfrac{1}{9}\dfrac{d}{{dx}}\left( {{y^2}} \right) = 0$
On using formula $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$ and using chain rule in the above equation for differentiation of y, we get-
$\Rightarrow \dfrac{1}{{{a^2}}}\left( {2{x^{2 - 1}}} \right) + \dfrac{1}{9}\left( {2{y^{2 - 1}}} \right)\dfrac{{dy}}{{dx}} = 0$
On solving we get,
$\Rightarrow \dfrac{1}{{{a^2}}}\left( {2x} \right) + \dfrac{1}{9}\left( {2y} \right)\dfrac{{dy}}{{dx}} = 0$ $\Rightarrow \dfrac{1}{{{a^2}}}\left( {2x} \right) + \dfrac{1}{{{b^2}}}\left( {2y} \right)\dfrac{{dy}}{{dx}} = 0$
On taking $2$ common we get,
$\Rightarrow 2\left[ {\dfrac{1}{{{a^2}}}\left( x \right) + \dfrac{1}{9}\left( y \right)\dfrac{{dy}}{{dx}}} \right] = 0$
$\Rightarrow \dfrac{1}{{{a^2}}}\left( x \right) + \dfrac{1}{9}\left( y \right)\dfrac{{dy}}{{dx}} = 0$
We can also write it as-
$\Rightarrow \dfrac{1}{{{a^2}}}\left( x \right) = - \dfrac{1}{9}\left( y \right)\dfrac{{dy}}{{dx}}$
On transferring x we get,
$\Rightarrow \dfrac{1}{{{a^2}}} = - \dfrac{1}{{9x}}\left( y \right)\dfrac{{dy}}{{dx}}$ -- (iii)
From eq. (ii) and eq. (iii) we get,
$\Rightarrow \dfrac{1}{{{x^2}}}\left( {1 - \dfrac{{{y^2}}}{9}} \right) = \dfrac{{ - y}}{{9x}}\dfrac{{dy}}{{dx}}$
On simplifying we get,
$\Rightarrow 1 - \dfrac{{{y^2}}}{9} = {x^2}\dfrac{{ - y}}{{9x}}\dfrac{{dy}}{{dx}}$
We can also write this equation as-
$\Rightarrow {x^2}\dfrac{{ - y}}{{9x}}\dfrac{{dy}}{{dx}} = 1 - \dfrac{{{y^2}}}{9}$
On adjusting the equation we get,
$\Rightarrow {x^2}\dfrac{{ - y}}{{9x}}\dfrac{{dy}}{{dx}} + \dfrac{{{y^2}}}{9} = 1$
$\Rightarrow x\dfrac{{ - y}}{9}\dfrac{{dy}}{{dx}} + \dfrac{{{y^2}}}{9} = 1$
Now we can also write $\dfrac{{dy}}{{dx}} = {y^{'}}$ then the equation becomes,
$\Rightarrow x\dfrac{{ - y}}{9}y' + \dfrac{{{y^2}}}{9} = 1$
On taking LCM we get,
$\Rightarrow \dfrac{{ - xyy' + {y^2}}}{9} = 1$
On transferring $9$ to the right side, we get
$\Rightarrow - xyy' + {y^2} = 9$
On multiplying by (-) ve sign both side we get,
$\Rightarrow xyy' - {y^2} = - 9$
$\Rightarrow xyy' - {y^2} + 9 = 0$

Hence, the correct answer is D.

Note:
To form a differential equation from the ellipse we first eliminate the constants then differentiate the equation. Here we used the points $\left( {0,3} \right)$ to put value of b in the equation .But we did not put a=$0$ because then the independent variable x will become undefined and the calculation will become complex.