The different equation of the family of parabolas with focus at origin and x-axis as axis is

y {(\frac{d y}{d x})}^{2} + 4 x \frac{d y}{d x} = 4 y

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- y {(\frac{d y}{d x})}^{2} = 2 x \frac{d y}{d x} - y

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

y {(\frac{d y}{d x})}^{2} + y = 2 x y \frac{d y}{d x}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

y {(\frac{d y}{d x})}^{2} + 2 x y \frac{d y}{d x} + y = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $- y {(\frac{d y}{d x})}^{2} = 2 x \frac{d y}{d x} - y$
General equation will be
$y^{2} = 4 a (a + x)$
Now differentiating , we get
$2 y \frac{d y}{d x} = 4 a$
$\Rightarrow y \frac{d y}{d x} = 2 a$
Now by putting this value in general equation, we get
$y^{2} = 2 y \frac{d y}{d x} (\frac{y}{2} \frac{d y}{d x} + x)$
$y^{2} = y^{2} (\frac{d y}{d x})^{2} + 2 y x \frac{d y}{d x}$
or $y^{2} = y^{2} y_{1}^{2} + 2 x y y_{1}$
$\to y = y y_{1}^{2} + 2 x y_{1}$
or $- y (\frac{d y}{d x})^{2} = 2 x \frac{d y}{d x} - y$

Suggest Corrections

Similar questions

Q. Differential equation of the family of parabolas whose vertex lie on the

x -

axis and focus as origin is

y {(\frac{d y}{d x})}^{2} + y = 2 x y \frac{d y}{d x}

Formation of the differential equation corresponding to the ellipse major axis 2a and minor axis 2b is:

The differential equation of family of curves $y^{2} = 4 a (x + a)$ is
a) $y^{2} = 4 \frac{d y}{d x} (x + \frac{d y}{d x})$
b) $2 y \frac{d y}{d x} = 4 a$
c) $y \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2} = 0$
d) $2 x \frac{d y}{d x} + y {(\frac{d y}{d x})}^{2} - y = 0$

{(\frac{d y}{d x})}^{2} - (x + y) \frac{d y}{d x} + x y = 0

Join BYJU'S Learning Program

The different equation of the family of parabolas with focus at origin and x-axis as axis is

The correct option is D −y(dydx)2=2xdydx−yGeneral equation will be y2=4a(a+x)Now differentiating , we get2ydydx=4a⇒ydydx=2aNow by putting this value in general equation, we gety2=2ydydx(y2dydx+x)y2=y2(dydx)2+2yxdydxor y2=y2y21+2xyy1→y=yy21+2xy1or −y(dydx)2=2xdydx−y