Question

help pleasee class-12

Answer 1

$\rm \to \green{{ \underline{given \div }}}$

$\rm \to \: curve \: = y \: = x {}^{3} + 2x + 6 \: \\ \rm \to \: perpendicular \: to \: the \: line \: = x \: + 14y + 4 = 0$

$\rm \to \: differentiation \: of \: \: curve \: \: w.r.t \: \frac{dy}{dx}$

$\rm \to \: \frac{dy}{dx} = 3 {x}^{2} + 2$

$\rm \to \: slope \: = x \: + 14y + 4 = 0 \\ \rm \to \: x \: + 4 = - 14y \\ \rm \to \: y \: = \frac{ - x}{14} - \frac{4}{14} \\ \rm \to \: slope \: = \frac{ - 1}{14}$

$\rm \to \: \frac{ - 1}{3 {x}^{2} + 2 } = \frac{ - 1}{14} \\ \\ \rm \to \: 3 {x}^{2} = 12 \\ \\ \rm \to \: x \: = + 2 \: \: and \: \: - 2$

when x = 2

=> y = (2) ³ + 2(2) +6

=> y = 18

when x = -2

=> y = (-2) ³ + 2(-2) + 6

=> y = -6

Therefore,

coordinate = ( 2,18) and (-2, -6)

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

For coordinate = ( 2,18)

=> ( y - 18 ) = 14 ( x - 2 )

=> y - 18 = 14x - 28

=> 14x - y - 10 = 0

=> 14x - y = 10

For coordinate = ( -2, -6)

=> ( y + 6 ) = 14 ( x + 2 )

=> y + 6 = 14x + 28

=> 14x - y + 22 = 0

=> 14x - y = -22

=> some related formula.

1) = equation of normal.

=> ( y - y1) = -1/m ( x - x1)

2) = equation of tangent

=> ( y - y1) = m ( x - x1)

3) = slope of equation

=> y2 - y1 / x2 - x1

Answer 2

Answer:

⭐ Solution ⭐

✏curve=y=x³ +2x+6

✏perpendiculartotheline

=x+14y+4=0

✏differentiation of curve w.r.t

= dx/dy

✏dx/dy =3x² +2

✏slope=x+14y+4=0

✏x+4=−14y

✏y= −x/14 − 4/14

✏slope= −1/14

✏-1/3x² +2−1 = -1/14

✏3x²=12

✏ x=+2and−2

✍when x = 2

=> y = (2)³+ 2(2) +6

=> y = 18

✍when x = -2

=> y = (-2) ³ + 2(-2) + 6

=> y = -6

➡Therefore,

coordinate = ( 2,18) and (-2, -6)

✏equation of tangent

=(y−y1 )=m(x−x 1 )

✏For coordinate = ( 2,18)

=> ( y - 18 ) = 14 ( x - 2 )

=> y - 18 = 14x - 28

=> 14x - y - 10 = 0

=> 14x - y = 10

✍For coordinate = ( -2, -6)

=> ( y + 6 ) = 14 ( x + 2 )

=> y + 6 = 14x + 28

=> 14x - y + 22 = 0

=> 14x - y = -22

Step-by-step explanation:

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