Question

Find the equation of the tangent and normal to the following curves at the indicated points.

(a) 4x² + 25y = 200 at (-5 , 2)

(b) x³ + xy² - 3x² + 4x + 5y + 20 = 0. at (1 , -1)
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Answer 1

Solution:-Finding the tangent and normal of the given terms,is not a very tough task

what we have to do,is to differentiate the term with respect to x and putting the coordinates to get the slope of tangent at that point

a)

$4 {x}^{2} + 25y = 200 \\ 4 {x}^{2} - 200 = - 25y \\ 200 - 4 {x}^{2} = 25y \\ 8 - \frac{4 {x}^{2} }{25} = y \\ \frac{dy}{dx} = 0 - \frac{8x}{25} \\ \frac{dy}{dx} = \frac{8x}{25} \\ now \: put \: the \: point \: at \: which \: \\ we \: have \: to \: find \: the \: tangent \\ slope \: of \: tangent = \frac{ - 5 \times 8}{25} = \frac{ - 8}{5} \\ eq \: of \: straight \: line = > y = mx + c \\ eqution \: of \: tangent = > y = \frac{ - 8}{5} x + c \\ 5y + 8x - 5c = 0 \\ putting \: x = - 5 \: and \: y = 2 \\ we \: get \\ c = - 6 \\ hence \: equation \: of \: tangent = 5y + 8x + 30 = 0 \\ normal \: is \: perpendicular \: to \: slope \: so \\ slope \: of \: normal = \frac{5}{8} \\ \: its \: equation = > y = \frac{5}{8}x + c \\ 8y - 5x - c = 0 \\ putting \: coordinate( - 5 \: 2) \: we \: get \\ 16 - 25 = c \\ c = - 9 \\ equation \: of \: normal = > 8 y - 5x + 9= 0$

b) we will use implicit rule of differenitaiation to take out dy/dx

$3 {x}^{2} + 1 \times {y}^{2} + x \times \frac{d {(y)}^{2} }{dy} \times \frac{dy}{dx} - 6x + 4 \\ + \frac{d(5y)}{dy} \times \frac{dy}{dx} + 0 = 0 \\ 3 {x}^{2} + {y}^{2} + 2y \times \frac{dy}{dx} - 6x + 4 + 5 \times \frac{dy}{dx} = 0 \\ \frac{dy}{dx} (2y + 5) = - 3 {x}^{2} - {y}^{2} + 6x + 4 \\ \frac{dy}{dx} = \frac{ - 3 {x}^{2} - {y}^{2} + 6x + 4}{2y + 5} \\ slope \: of \: tangent \: at \: (1 \: - 1) \: is \: = \frac{ - 3 - 1 + 6 + 4}{ - 2 + 5} \\ slope \: of \: tangent = \frac{6}{3} = 2 \\ equation \: of \: tangent = > y = 2x + c \\ putting \: x = 1 \: and \: y = - 1 \: in \: eq \: we \: get \\ - 1 - 2 = c \\ c = - 3 \\ equation \: of \: tangent = > 2x - y - 3 = 0 \\ slope \: of \: normal = \frac{ - 1}{2} \\ equation \: of \: normal = > y = \frac{ - 1}{2}x + c \\ putting \: the \: coordinates \: we \: get \\ c = - 1 + \frac{1}{2} = \frac{ - 1}{2} \\ 2y = - x - 1 \\ equation \: of \: normal = > x + 1 + 2y = 0$

Answer 2

Answer:

Refer to the attachment