Question

find equation to the tangent to x²+y² = 8 at (2,2)​

Answer 1

$\large\underline{\sf{Solution-}}$

Given curve is

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = 8$

On differentiating w. r. t. x, we get

$\rm :\longmapsto\: \dfrac{d}{dx}({x}^{2} + {y}^{2} )=\dfrac{d}{dx} 8$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \: }} \: \: and \: \: \boxed{ \tt{ \: \dfrac{d}{dx}k \: = \: 0 \: }}$

So, using these, we get

$\rm :\longmapsto\:2x + 2y\dfrac{dy}{dx} = 0$

$\rm :\longmapsto\:x + y\dfrac{dy}{dx} = 0$

$\rm :\longmapsto\: y\dfrac{dy}{dx} = - \: x$

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

We know,

Let y = f(x) be any curve, then slope of tangent at any point (a, b) is given by

$\boxed{ \tt{ \: Slope \: of \: tangent, \: m \: = \: \bigg(\dfrac{dy}{dx}\bigg) _{(a,b)} \: }}$

Slope of tangent to the given curve at (2, 2) is

$\rm :\longmapsto\: Slope \: of \: tangent, \: m \: = \: \bigg(\dfrac{dy}{dx}\bigg) _{(2,2)} \:$

$\rm :\longmapsto\: Slope \: of \: tangent, \: m \: = \: - \: \dfrac{2}{2}$

$\red{\rm :\longmapsto\: Slope \: of \: tangent, \: m \: = \: - \: 1 }$

So, equation of tangent which passes through the point (2, 2) having slope - 1, is

$\rm :\longmapsto\:y - 2 = - 1(x - 2)$

$\rm :\longmapsto\:y - 2 = - x + 2$

$\rm :\longmapsto\:y + x =2 + 2$

$\bf\implies \:\boxed{ \tt{ \: x + y \: = \: 4 \: }}$

Explore more

1. Let y = f(x) be any curve, then line which touches the curve y = f(x) exactly at one point say P is called tangent and that very point P, if we draw a perpendicular on tangent, that line is called normal to the curve at P.

2. If tangent is parallel to x - axis, its slope is 0.

3. If tangent is parallel to y - axis, its slope is not defined

4. Two lines having slope M and m are parallel, iff M = m

5. If two lines having slope M and m are perpendicular, iff Mm = - 1.

$\red{\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}}$

Answer 2

Answer:

$\large\underline{\sf{Solution-}}$

Given curve is

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = 8$

On differentiating w. r. t. x, we get

$\rm :\longmapsto\: \dfrac{d}{dx}({x}^{2} + {y}^{2} )=\dfrac{d}{dx} 8$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \: }} \: \: and \: \: \boxed{ \tt{ \: \dfrac{d}{dx}k \: = \: 0 \: }}$

So, using these, we get

$\rm :\longmapsto\:2x + 2y\dfrac{dy}{dx} = 0$

$\rm :\longmapsto\:x + y\dfrac{dy}{dx} = 0$

$\rm :\longmapsto\: y\dfrac{dy}{dx} = - \: x$

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

We know,

Let y = f(x) be any curve, then slope of tangent at any point (a, b) is given by

$\boxed{ \tt{ \: Slope \: of \: tangent, \: m \: = \: \bigg(\dfrac{dy}{dx}\bigg) _{(a,b)} \: }}$

Slope of tangent to the given curve at (2, 2) is

$\rm :\longmapsto\: Slope \: of \: tangent, \: m \: = \: \bigg(\dfrac{dy}{dx}\bigg) _{(2,2)} \:$

$\rm :\longmapsto\: Slope \: of \: tangent, \: m \: = \: - \: \dfrac{2}{2}$

$\red{\rm :\longmapsto\: Slope \: of \: tangent, \: m \: = \: - \: 1 }$

So, equation of tangent which passes through the point (2, 2) having slope - 1, is

$\rm :\longmapsto\:y - 2 = - 1(x - 2)$

$\rm :\longmapsto\:y - 2 = - x + 2$

$\rm :\longmapsto\:y + x =2 + 2$

$\bf\implies \:\boxed{ \tt{ \: x + y \: = \: 4 \: }}$

Explore more

1. Let y = f(x) be any curve, then line which touches the curve y = f(x) exactly at one point say P is called tangent and that very point P, if we draw a perpendicular on tangent, that line is called normal to the curve at P.

2. If tangent is parallel to x - axis, its slope is 0.

3. If tangent is parallel to y - axis, its slope is not defined

4. Two lines having slope M and m are parallel, iff M = m

5. If two lines having slope M and m are perpendicular, iff Mm = - 1.

$\red{\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}}$