Question

Find the points on the curve y = x3 – 2x2 - x
where the tangents are parallel to 3x-y+1=0.
Find the equations of the tangents to the
curve x2 + y2 – 2x - 4y + 1= 0 which are
parallel to the X-axis.
'ind the equations of the normals to the
irve 3x2 - y2 = 8, which are parallel to the​

Answer 1

hope it helps you mate.

Answer 2

1) = Find the point on the curve y = x³ - 2x² - x

where the tangent are parallel to 3x - y + 1 = 0

$\sf \to \: 3x - y + 1 = 0 \\ \\ \sf \to \: y \: = 3x + 1 = 0 \\ \\ \sf \to \: \dfrac{dy}{dx} = 3 = slope \\ \\ \sf \to \: \dfrac{dy}{dx} = 3 {x}^{2} - 4x - 1 = 0 \\ \\ \sf \to \: \frac{dy}{dx} = 3 {x}^{2} - 4x - 1 = 3 \\ \\ \sf \to \: \frac{dy}{dx} = 3 {x}^{2} - 4x - 4 = 0 \\ \\ \sf \to \: 3 {x}^{2} - 6x + 2x - 4 = 0 \\ \\ \sf \to \: 3x(x - 2) + 2(x - 2) = 0 \\ \\ \sf \to \: (3x + 2)(x - 2) = 0 \\ \\ \sf \to \: x \: = \frac{ - 2}{3} \: and \: x \: = 2$

$\sf \to \: put \: the \: value \: of \: x \: in \: equation \\ \\ \sf \to \: when \: x \: = \frac{ - 2}{3} = y = \frac{ ( - 2) {}^{3} }{(3) {}^{3} } - 2 \frac{( - 2) {}^{2} }{(3) {}^{2} } - \frac{( - 2)}{(3)} = 0 \\ \\ \sf \to \: y \: = \dfrac{ - 14}{27} \\ \\ \sf \to \: when \: x \: = 2 \\ \\ \sf \to \: y \: = (2) {}^{3} - 2(2) {}^{2} - 2 \\ \\ \sf \to \: y \: = - 2 \\ \\ \sf \to \: \green{{ \underline{point \: on \: the \: curve \: = ( \frac{ - 2}{3} \: , \frac{ - 14}{27}) \: \: and \: \: \: (2 \:, \: - 2) }}}$

2) = find the equation of tangents to the curve

=> x² + y² - 2x - 4y + 1 = 0 which are parallel

to the x-axis axis

$\sf \to \: parallel \: to \: x \: axis \: \\ \\ \sf \to \: differentiate \: w.r.t \: y \\ \\ \sf \to \: equation \: = {x}^{2} + {y}^{2} - 2x - 4y + 1 = 0 \\ \\ \sf \to \: 2x \dfrac{dy}{dx} + 2y - 2 \frac{dy}{dx} - 4 = 0 \\ \\ \sf \to \: (2x - 2) \dfrac{dy}{dx} = 4 - 2y \\ \\ \sf \to \: \dfrac{dy}{dx} = \dfrac{2x - 2}{4 - 2y} \\ \\ \sf \to \: 2x - 2 = 0 \\ \\ \sf \to \: x \: = 1$

$\sf \to \: put \: the \: value \: of \: x \: = 1 \: in \: equation \: \\ \\ \sf \to \: y = (1) {}^{2} + {y}^{2} - 2(1) - 4y + 1 = 0 \\ \\ \sf \to \: y = 1 + {y}^{2} - 2 - 4y + 1 = 0 \\ \\ \sf \to \: {y}^{2} - 4y = 0 \\ \\ \sf \to \: y(y - 4) = 0 \\ \\ \sf \to \: y \: = 0 \: \: \: and \: \: y \: = 4 \\ \\ \sf \to points \: are \: (1 \: 0) \: and \: (1 \: 4) \\ \\ \sf \to \: equation \: of \: tangent \: (y - y_{1}) = \frac{dy}{dx}(x - x_{1}) \\ \\ \sf \to \: (y - 0) = 0 \\ \\ \sf \to \: y \: = 0 \\ \\ \sf \to \: (y - 1) = 0 \\ \\ \sf \to \: y \: = 1$

3) = Find the equation of the normal lines to

the curve 3x² - y = 8 which are parallel to

the line x + 3y = 4

$\sf \to \: equation \: 3 {x}^{2} - {y}^{2} = 8 \\ \\ \sf \to \: differentiate \: w.r.t \: x \\ \\ \sf \to \: 6x - 2y \frac{dy}{dx} = 0 \\ \\ \sf \to \: slope \: of \: tangent \: = \frac{3x}{y} \\ \\ \sf \to \: slope \: of \: normal \: = \frac{ - y}{3x} \\ \\ \sf \to \: slope \: = \frac{ - 1}{3} \\ \\ \sf \to \: \frac{ - y}{3x} = \frac{ - 1}{3} \\ \\ \sf \to \: y \: = x \\ \\ \sf \to \: 3 {x}^{2} - {x}^{2} = 8 \\ \\ \sf \to \: 2 {x}^{2} = 8 \\ \\ \sf \to \: {x}^{2} = 4 \\ \\ \sf \to \: x \: = 2 \: \: and \: x \: = - 2 \\ \\ \sf \\ \to \: when \: x \: = 2 \: and \: y \: = 2 \\ \\ \sf \to \: when \: x \: = - 2 \: and \: y \: = - 2 \\ \\ \sf \to \: equation \: of \: normal$

=> ( y - 2 ) = -1/3 ( x - 2 )

=> 3y - 6 = -x + 2

=> 3y + x = 8

=> ( y + 2 ) = -1/3 ( x + 2 )

=> 3y + 6 = - x - 2

=> 3y + x = -8