Question

find equation of tangent to curve x²+y²=5 where tangent is parallel to line 2x-y+1=0​

Answer 1

Step-by-step explanation:

We have,

$\boxed{ \tt{ \green{ Given \: \: curve \colon\: \: \pink{ {x}^{2} + {y}^{2} = 5 } }}}$

Differentiating it w.r.t. x,

$\sf{2x + 2y \dfrac{dy}{dx} = 0 }$

$\sf{ \implies \dfrac{dy}{dx} = - \dfrac{x}{y} }$

$\sf{ \implies \blue{ \dfrac{dy}{dx} \bigg|_{(h , k)}= - \dfrac{h}{k}} }$

Now,

the tangent is parallel to $\boxed{\purple{\rm{2x-y+1=0}}}$

Slope of the above line is 2

So,

$\sf{ - \dfrac{h}{k} = 2}$

$\sf{ \implies h = - 2k}$

(h,k) lies on the curve, to find the coordinates of this point put (h,k) on the curve,

So,

$\sf{ {h}^{2} + {k}^{2} = 5}$

$\sf{ \implies {( - 2k)}^{2} + {k}^{2} = 5}$

$\sf{ \implies 4{k}^{2} + {k}^{2} = 5}$

$\sf{ \implies 5{k}^{2} = 5}$

$\sf{ \implies {k}^{2} = 1}$

$\sf{ \implies k = \pm 1}$

So,

$\sf{h = \mp 2}$

So required coordinates of points , where, tangent is parallel to the given line are

$\sf{\color{cyan}(\pm1,\mp2)}$

$\boxed{ \large{ \rm{ \bold{ \red{ \bull \: \: Equation \: \: of \: \: tangent \colon }}}}}$

$\sf{ \{y - ( \pm 1) \} = 2 \{x - ( \mp2) \} }$

$\sf{ \implies \: y \mp 1 = 2 (x \pm2) }$

$\sf{ \implies \: y \mp 1 = 2x \pm4 }$

$\sf{ \implies \: y - 2x= \pm4 \pm1 }$

$\sf{ \implies \: y - 2x= \pm5 }$