If $a x^{2} + 2 h x y + b y^{2} = 1$ , then $y_{2} =$

\frac{h^{2} - a b}{h x + b y}

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

\frac{h^{2} - a b}{(h x + b y)^{3}}

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

\frac{h^{2} + a b}{(h x + b y)^{3}}

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

\frac{h^{2} + a b}{h x + b y}

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

Open in App

Solution

The correct option is B $\frac{h^{2} - a b}{(h x + b y)^{3}}$
$G i v e n t h a t a x^{2} + 2 h x y + b y^{2} = 1 T h e n d i f f e r e n t i a t e t h e e q u a t i o n w . r . t x ⟹ 2 a x + 2 h (x y_{2} + y \cdot 1) + 2 b y y_{1} = 0 a x + h x y_{1} + h y + b y y_{1} = 0 (h x + b y) y_{1} = - a x - h y y_{1} = - \frac{(a x + h y)}{(h x + b y)} . . . . . (1) f u r t h e r d i f f e r e n t i a t i n g e q u a t i o n (1) w . r . t x y_{2} = - \frac{[(h x + b y) (a + h y_{1}) - (a x + h y) (h + b y_{1})]}{{(h x + b y)}^{2}} B y s i m p l i f y i n g f u t h e r y_{2} = \frac{h^{2} - a b}{{(h x + b y)}^{3}} H e n c e O P T I O N B i s c o r r e c t .$

Suggest Corrections

Similar questions

(1) \frac{x^{2}}{b c - f^{2}} = \frac{y^{2}}{c a - g^{2}} = \frac{z^{2}}{a b - h^{2}} .

(2) (b c - f^{2}) (c a - g^{2}) (a b - h^{2}) = (f g - c h) (g h - a f) (h f - b g) .

a x + h y + g = 0

h x + b y + f = 0

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c + t = 0

\frac{a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2}}{h^{2} - a b} = 8,

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

(\frac{b g - h f}{h^{2} - a b}, \frac{a f - g h}{h^{2} - a b})

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

(\sqrt{(\frac{f^{2} - b c}{h^{2} - a b})}, \sqrt{(\frac{g^{2} - a c}{h^{2} - a b})})

P (h, k)

The angle between the lines represented by the equation $a x^{2} + 2 h x y + b y^{2} = 0$ is given by

Join BYJU'S Learning Program