# Find the differential equations of the following family of curves.

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Bingo
Find the differential equations of the following family of curves.

Differential Equations
Find the differential equations of the following family of curves.

1. Parabolas with axis parallel to the y – axis with distance vertex to focus fixed as a.
2. Parabolas with axis parallel to the x – axis with distance vertex to focus fixed as a.
3. All ellipses with center at the origin and axes on the coordinate axes.
4. Family of cardioids.
5. Family of 3 – leaf roses.

Bingo

1

Jhun Vert

(1) Upwards and downwards parabolas with latus rectum equal to 4a.
$y - k = \pm 4a(x - h)^2$

$y' = \pm 8a(x - h)$

$y'' = \pm 8a$

Bingo

y−k=±4a(x−h)2 is this the equation for parabolas?
isn't it 4a(y-k)=(x-h)2 ?

Jhun Vert

I made a mistake in there, it should be (x - h)2 = ±4a(y - k). The solution for (1) should go this way:
$(x - h)^2 = \pm 4a(y - k)$

$2(x - h) = \pm 4ay'$

$2 = \pm 4ay''$

$y'' = \pm \frac{1}{2a}$

Bingo

y−k=±4a(x−h)2 is this the equation for parabolas?
isn't it 4a(y-k)=(x-h)2 ?

Jhun Vert

Bingo

so for number 2,
x-h=±4a(y-k)2
x'=±8a(y-k)
x''=±8a
am i wrong?

Jhun Vert

It is better to express your answer in terms of y' rather than x'. Although x' will do and simpler.
$(y − k)^2 = \pm 4a(x − h)$

$2(y − k)y' = \pm 4a$

$y' = \dfrac{\pm 2a}{y - k}$

Note: $y - k = \dfrac{y'}{\pm 2a}$

$y'' = \dfrac{\mp 2a \, y'}{(y - k)^2}$

$y'' = \dfrac{\mp 2a \, y'}{\left( \dfrac{y'}{\pm 2a} \right)^2}$

$y'' = \dfrac{\mp 8a}{y'}$

$y'' \, y' = \pm 8a$

ChaCha Ingal Cortez

our prof. gave the same question but the ans. he gave is different from your ans. in no 2, his answer is (y')^3 + 2ay"=0 are they the same with your answer sir ?

Jhun Vert

They are different. I made a mistake in $y - k$ of the above solution.
$(y − k)^2 = \pm 4a(x − h)$

$2(y - k) \, y' = \pm 4a$

$y' = \dfrac{\pm 2a}{y - k}$

$y - k = \dfrac{\pm 2a}{y'}$

$y'' = \dfrac{\mp 2a \, y'}{(y - k)^2}$

$y'' = \dfrac{\mp 2a \, y'}{\left( \dfrac{\pm 2a}{y'} \right)^2}$

$y'' = \dfrac{\mp 2a \, y'}{\dfrac{4a^2}{(y')^2}}$

$y'' = \dfrac{\mp (y')^3}{2a}$

$2a \, y'' = \mp (y')^3$

$\pm (y')^3 + 2a \, y'' = 0$

Bingo

Thank You so much sir.

Bingo

sir, do you know the standard or general equations for the last 3 problems?
what equations should i use?

Shandy

I think you are from ADZU haha

Alegnaaa (guest)

Good evening, sir. Do you have the solution for the last three problems? Thank you!

Ancha (guest)

Sir do you know how to eliminate arbitrary constant here in this equation?
(y-33)²=4a(x-h)?

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