Question

Find the slope of tangent to the curve x2+y2=25 at point (-3,4)​

Answer 1

EXPLANATION.

→ Slope of tangent to the curve → x² + y² = 25.

at point → ( -3,4).

→ Differentiate w.r.t. x, we get,

→ 2x + 2y.dy/dx = 0

→ 2x = - 2y.dy/dx.

→ dy/dx = -x/y.

→ Put the value of x and y

→ dy/dx = -(-3)/4.

→ dy/dx = 3/4.

→ Slope of the tangent = 3/4.

→ More information.

→ Equation of the tangent.

$\sf : \implies \: (y - y_{1}) = m(x - x_{1})$

→ ( y - 4) = 3/4 ( x - ( -3))

→ 4 ( y - 4 ) = 3 ( x + 3 )

→ 4y - 16 = 3x + 9

→ 4y - 3x = 25

→ Equation of Normal.

$\sf : \implies \: (y - y_{1}) = \dfrac{ - 1}{m} (x - x_{1})$

→ ( y - 4 ) = -4/3 ( x + 3 )

→ 3( y - 4 ) = -4 ( x + 3 )

→ 3y - 12 = -4x - 12

→ 3y + 4x = 0

Answer 2

Given:-

• x2+y2=25 at point (-3,4)

To find:-

• slope of tangent to the curve x2+y2=25 at point (-3,4).

Solution:-

⟹ $\sf{\dfrac{2x + 2y.dy}{dx = 0}}$

⟹ $\sf{\dfrac{2x = - 2y.dy}{dx}}$

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

★ Put the value of x and y

$\sf{\dfrac{dy}{dx}}$ = $\sf{\dfrac{-(-3)}{4}}$

$\sf{\dfrac{dy}{dx}}$ = $\sf{\dfrac{3}{4}}$

•°• Slope of tangent is $\sf{\dfrac{3}{4}}$