Find the equations of the normals to the curve 3x2-y2=8 parallel to the line x+3y=4 - Maths - Application of Derivatives

Question

Find the equations of the normals to the curve
3x2-y2=8 parallel to the line x+3y=4

Gursheen kaur. · Accepted Answer

Let the point be $P (x_{1}, y_{1})$

$3 x^{2} - y^{2} = 8 6 x - 2 y \frac{d y}{d x} = 0 \frac{d y}{d x} = \frac{3 x}{y} t h e s l o p e o f t h e \tan g e n t i s m_{1} = {(\frac{d y}{d x})}_{p} = \frac{3 x_{1}}{y_{1}} ............... (i) t h e s l o p e o f l i n e x + 3 y = 4 i s m_{2} = - \frac{1}{3} ........................... (i i)$

As the line is parallel to the normal , it is perpendicular to the tangent.

$m_{1} * m_{2} = - 1 (\frac{3 x_{1}}{y_{1}}) (- \frac{1}{3}) = - 1 x_{1} = y_{1} a s, P (x_{1}, y_{1}) l i e s o n t h e c u r v e, w e g e t, 3 {x_{1}}^{2} - {y_{1}}^{2} = 8 3 {x_{1}}^{2} - {x_{1}}^{2} = 8 2 {x_{1}}^{2} = 8 {x_{1}}^{2} = 4 x_{1} = \pm 2 & y_{1} = \pm 2 t h e r e f o r e, t h e r e a r e t w o p t s . (2, 2) & (- 2, - 2) . t h e e q u a t i o n o f t h e n o r m a l a t (2, 2) i s y - 2 = - \frac{1}{3} (x - 2) x + 3 y = 8 a n d t h e e q u a t i o n o f t h e n o r m a l a t (- 2, - 2) i s y + 2 = \frac{- 1}{3} (x + 2) x + 3 y = - 8$

Find the equations of the normals to the curve 3x2-y2=8 parallel to the line x+3y=4.

Find the equations of the normals to the curve 3x²-y²=8 parallel to the line x+3y=4.