Find the orthogonal trajectory of the circle $x^{2} + y^{2} - a y = 0$

x^{2} + y^{2} = 4 c x y

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

x^{2} + y^{2} = c x

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

x^{2} - y^{2} = c x^{2}

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

x^{2} + y^{2} = c y

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

Open in App

Solution

The correct option is B $x^{2} + y^{2} = c x$
$\begin{matrix} x^{2} + y^{2} - a y = 0 2 x + 2 y \frac{d y}{d x} = a \frac{d y}{d x} = 0 2 x + 2 y . \frac{d y}{d x} = a \frac{d y}{d x} 2 x \frac{d x}{d y} + 2 y = a N o w, x^{2} + y^{2} - y (2 x . \frac{d x}{d y} + 24) = 0 x^{2} + y^{2} - 2 x y \frac{d y}{d x} - 2 y^{2} = 0 x^{2} - y^{2} = 2 x y \frac{d y}{d x} F o r o t h o g o n a l t r a j e c t o r y \frac{d y}{d x} = - \frac{d x}{d y} ∴ y^{2} - x^{2} = 2 x y . \frac{d y}{d x} \frac{d y}{d x} = \frac{y^{2} - x^{2}}{2 x y} ∴ x^{2} + y^{2} = c x H e n c e, t h e o p t i o n B i s t h e c o r r e c t a n s w e r . \end{matrix}$

Suggest Corrections

Similar questions

The orthogonal trajectory of y^{2} = 4 ax is

Find the orthogonal trajectory of $x^{2} + y^{2} = c$

x^{2} + y^{2} = c

x^{2} - y^{2} = c

x^{2} + y^{2} - 8 x - 6 y + 21 = 0

x^{2} + y^{2} + a y - 15 = 0

a

Join BYJU'S Learning Program