Question

find equation of tangent and equation of normal at point (-1,2) for the curve y⁴-4x⁴-6xy=0​

Answer 1

EXPLANATION.

→ Equation of tangent and normal at point

( -1,2) for curve → y⁴ - 4x⁴ - 6xy = 0.

→ Differentiate w.r.t x we get,

→ 4y³.dy/dx - 16x³ - 6x.dy/dx - 6y X 1 = 0

→ ( 4y³ - 6x ) dy/dx = 16x³ + 6y

→ dy/dx = 16x³ + 6y / 4y³ - 6x

→ dy/dx = 8x³ + 3y / 2y³ - 3x

→ dy/dx at point ( -1,2)

→ dy/dx = 8(-1)³ + 3(2) / 2(2)³ - 3(-1)

→ dy/dx = -8 + 6 / 16 + 3

→ dy/dx = -2/19

→ Slope of equation = -2/19.

$\sf : \implies \: equation \: of \: tangent \implies \: (y - y_{1}) = m(x - x_{1})$

→ ( y - 2 ) = -2/19 ( x - (-1))

→ 19 ( y - 2 ) = -2 ( x + 1 )

→ 19y - 38 = -2x - 2

→ 19y + 2x = 36 = equation of tangent.

$\sf : \implies \: equation \: of \: normal \implies \: (y - y_{1}) = \dfrac{ - 1}{m} (x - x_{1})$

→ ( y - 2 ) = 19/2 ( x + 1 )

→ 2 ( y - 2 ) = 19 ( x + 1 )

→ 2y - 4 = 19x + 19

→ 2y - 19x = 23 = equation of normal.