Question

$\huge{He¥ MaTe!!!}$

☆☞ Here is ur Question ☜☆

Que - Show that the family y^{2} =4a(x+a) is self orthogonal.

NO SPAM PLEASE !​

Answer 1

Step-by-step explanation:

Given: y² = 4a(x + a)   --------- (i)

On Differentiating with respect to x, we get

⇒ 2y (dy/dx) = 4a

⇒ y (dy/dx) = 4a/2

⇒ y (dy/dx) = 2a

⇒ a = (1/2) y. dy/dx

Substitute value of a in (i), we get

$=>y^2 = 4*\frac{1}{2} *y*\frac{dy}{dx}(x+\frac{1}{2}y*\frac{dy}{dx})$

$=>y^2=2xy.\frac{dy}{dx}(x+\frac{1}{2}y*\frac{dy}{dx})$

$=>y^2=2xy.\frac{dy}{dx}+y^2(\frac{dy}{dx})^2$    ----- (ii)

This is the Differential Equation of the given family of parabolas.

On Replacing (dy/dx) by (-dx/dy) in (ii), we get

$=>y^2=2xy(-\frac{dx}{dy}) + y^2(-\frac{dx}{dy})^2$

$=>y^2=-2xy\frac{dy}{dx}+y^2(\frac{dx}{dy})^2$

$=>y^2(\frac{dy}{dx})^2+2xy.\frac{dy}{dx}=y^2$

Which is same as Equation (ii).

Thus, Differential Equation of the family of parabolas given by the (i), (ii) and the differential equation of its orthogonal trajectories are same.

Hence the given family is "Self-orthogonal".

Hope it helps!

Answer 2

Differentiate both sides with respect to x:

2y dy/dx = 4a

==> dy/dx = 2a/y.

To eliminate a:

y^2 = 4ax + 4a^2

==> x^2 + y^2 = x^2 + 4ax + 4a^2 = (x + 2a)^2

==> √(x^2 + y^2) = x + 2a

==> a = (1/2) [√(x^2 + y^2) - x].

Hence, dy/dx = 2a/y = [√(x^2 + y^2) - x]/y.

---------------------

So, the differential equation for the orthogonal trajectories is simply the negative reciprocal of dy/dx (found on the previous line):

==> dy/dx = -y/[√(x^2 + y^2) - x].

Now, we solve for y:

[√(x^2 + y^2) - x] dy = -y dx

==> -√((x/y)^2 + 1) + x/y = dx/dy.

Since this is homogeneous, let v = x/y.

So, x = yv ==> dx/dy = v + y dv/dy.

Now, the DE transforms to

-√(v^2 + 1) + v = v + y dv/dy

==> dv/√(v^2 + 1) = -dy/y

Integrate both sides:

ln(√(v^2 + 1) + v) = -ln y + A

==> √(v^2 + 1) + v = Cy^(-1), where C = e^A

==> √((x/y)^2 + 1) + x/y = C/y

==> √(x^2 + y^2) + x = C.

-------------------

P.S.: Regarding ∫ dv/√(v^2 + 1)...

Letting v = tan t yields

∫ (sec^2(t) dt)/sec t

= ∫ sec t dt

= ln |sec t + tan t| + C

= ln |√(v^2 + 1) + v| + C, via tan t = v/1 and 'sohcahtoa'