Question

find where the tangent is parallel to x axis for the curve x^3+y^3=a^3 also find equation of the tangent at that point​

Answer 1

EXPLANATION.

$\sf : \implies \: equation \: of \: curve \: = {x}^{3} + {y}^{3} = {a}^{3} \\ \\ \sf : \implies \: the \: tangent \: is \: parallel \: to \: x \: axis$

$\sf : \implies \: differentiate \: w.r.t \: \: \: x \\ \\ \sf : \implies \: 3 {x}^{2} + 3 {y}^{2} \frac{dy}{dx} = 0 \\ \\ \sf : \implies \: \frac{dy}{dx} = \frac{ - 3 {x}^{2} }{3 {y}^{2} } \\ \\ \sf : \implies \: \frac{dy}{dx} = \frac{ - {x}^{2} }{ {y}^{2} } \\ \\ \sf : \implies \: let \: the \: point \: be \: (h ,k)$

$\sf : \implies \: tangent \: is \: parallel \: to \: x \: axis \\ \\ \sf : \implies \: \frac{dy}{dx} = 0 \\ \\ \sf : \implies \: \frac{ - {h}^{2} }{ {k}^{2} } = 0 \\ \\ \sf : \implies \: - {h}^{2} = 0 \\ \\ \sf : \implies \: h \: \: = 0$

$\sf : \implies \: put \: the \: value \: of \: h \: = 0 \: in \: equation \\ \\ \sf : \implies \: 0 + {k}^{3} = {a}^{3} \\ \\ \sf : \implies \: k \: = a \\ \\ \sf : \implies \: points \: are \: = (0 ,a)$

$\sf : \implies \: equation \: of \: tangent \: = (y - y_{1}) = m(x - x_{1 }) \\ \\ \sf : \implies \: slope \: = 0 \\ \\ \sf : \implies \: (y - a) = 0(x - 0) \\ \\ \sf : \implies \: y \: = a \: \: \: equation \: of \: tangent$

Answer 2

$\sf{\gray{\underbrace{\red{ANSWER}}}}$

$\rm\huge{x}^{3} + {y}^{3} = {a}^{3}$

$\rm the \: tangent \: is \: parllel \: to \: x \: axis$

$\huge\rm{ {3x}^{2} + {3y}^{2} \: \: \frac{dy}{dx \: \: } = 0}$

$\huge\rm \: \frac{dy}{dx} = \frac{ { - 3x}^{2} }{ {3y}^{2} }$

$\huge \rm \: \frac{dy}{dx} = \frac{ { - x}^{2} }{ - {y}^{2} }$

$\rm \: let \: the \: point \: be \: (h \: k)$

$\rm \: tangent \: is \: parllel \: to \: x \: axis$

$\huge\rm \: \frac{dy}{dx} = 0$

$\huge\rm \frac{ { - h}^{2} }{{k}^{2} } = 0$

$\huge\rm \: h = 0$

$\rm \: put \: value \: of \: h \: = 0 \: in \: equation$

$\huge\rm \: 0 + {k}^{3 } = {a}^{2}$

$\huge\rm \: k \: = a$

$\rm \: points \: are \: = (0 \: a)$

$\rm \: equation \: of \: tangent \: = (y - y1)</p><p>$

$\huge\rm \: = m(x - x1)$

$\huge\rm \: slope = 0$

$\huge\rm \: (y - a) \: = \: 0(x - 0)$

$\rm \: y = a \: \: equation \: of \: tangent$