Question

Find the equation of all lines having slope 2 and being tangent to the curve y+2/x-3=0

Answer 1

Answer:

the answer is -2

Step-by-step explanation:

y+2/×-3=0

y+2=0

y=0-2

y=2

Hope it is useful.... Thanks

Answer 2

Giνєи —

Equation of curve $\sf{=y+\dfrac{2}{x-3}=0}$

Slope of lines = 2

Tσ Fiиd —

The equation of all lines being tangent to the given curve .

Sσℓυтiσи –

$\qquad\rm{=y+\dfrac{2}{x-3}=0}$

Differentiating both sides w.r.t x

$\qquad\rm{ \longrightarrow \dfrac{dy}{dx}+\dfrac{d}{dx}\left(\dfrac{2}{x-3}\right)=0}$

$\qquad\rm{ \longrightarrow \dfrac{dy}{dx}=-\dfrac{d}{dx}\left(\dfrac{2}{x-3}\right)}$

$\qquad\rm{ \longrightarrow \dfrac{dy}{dx}=-2\dfrac{d}{dx}(x-3)^{-1}}$

$\qquad\rm{ \longrightarrow \dfrac{dy}{dx}=-2\times -(x-3)^{-1-1}}$

$\qquad\rm{ \longrightarrow \dfrac{dy}{dx}=2(x-3)^{-2}}$

$\qquad\rm{ \longrightarrow \dfrac{dy}{dx}=\dfrac{2}{(x-3)^{2}}}$

$\qquad\rm{ \longrightarrow \dfrac{\cancel{2}}{(x-3)^2}=\cancel{2}\:\:\:\:\:\:\:\:\:\:\big\lgroup Given \:slope : \dfrac{dy}{dx}=2\big\rgroup}$

$\qquad\rm{ \longrightarrow \dfrac{1}{(x-3)^2}=1}$

$\qquad\rm{ \longrightarrow (x-3)^2 = 1}$

$\qquad\rm{ \longrightarrow x-3 = ±1}$

ㅤ

$\qquad \boxed{\begin {array}{c |c}\sf x - 3 = 1 &\sf x - 3 = - 1 \\\\\sf x = 1 + 3&\sf x = - 1 + 3 \\\\ \sf x = 4&\sf x = 2 \end{array}}$

❍ So , x = 4 and x = 2

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➨ Finding value of y

ㅤ

$\begin {array}{c |c} \underline{\large\bf{If \: x = 2}}& \underline{\large\bf{If \: x = 4}} \\\\ \sf y = \dfrac{ - 2}{x - 3} & \sf \: \: y = \dfrac{ - 2}{x - 3 } \\ \\ \sf y = \dfrac{ - 2}{2 - 3}& \sf \: y = \dfrac{ - 2}{4 - 3} \\\\ \sf \: y = \cancel{ \dfrac{ - 2}{ - 1}} & \sf \: y = \dfrac{ - 2}{1} \\\\ \sf \: y = 2& \sf \: y = - 2 \\\\ \rm Thus \: point \: is \: (2,2)& \rm \: Thus \: point \: is \: (4,-2) \end{array}$

ㅤ

Thus ,there are 2 tangents to the curve with slope 2 and passing through points ( 2,2 ) and ( 4,-2 )

✰ We know that :

Equation of line at $\sf{x_{1} ,y_{1}}$ and having slope m is :

$\qquad\underline{\boxed{\large\rm\purple{ \leadsto y-y_{1}=m(x-x_{1}) }}}$

$\boxed{\begin{array}{c |c}\underline{\bf{equation \: of \: tangent \: through \: (2,2)}}&\underline{\bf{equation \: of \: tangent \: through \: (4, - 2)}}\\\\ \\ \sf y - 2 = 2(x - 2)&\sf y - ( - 2) = 2(x - 8) \\\\\\\bold{ y - 2x + 2 = 0} &\bold{ y - 2x + 10 = 0}\end {array}}$

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