Question

Find an equation of the tangent line to the hyperbola y=3/X at the point (3,1)

Answer 1

Answer:

Let's find the derivative of that hyperbola in order to find a slope formula to help us with this equation. We already have an x and y value. The derivative is found this way: y'= \frac{x(0)-3(1)}{x^2}y

′

=

x

2

x(0)−3(1)

so y'=- \frac{3}{x^2}y

′

=−

x

2

3

. The derivative supplies us with the slope formula we need to write the equation. Sub in the x value of 3 to find what the slope is: y'=- \frac{3}{3^2}=- \frac{3}{9}=- \frac{1}{3}y

′

=−

3

2

3

=−

9

3

=−

3

1

. So in our slope-intercept equation, x = 3, y = 1, and m = -1/3. Use these values to solve for b. 1=- \frac{1}{3}(3)+b1=−

3

1

(3)+b so b = 2. The equation, then, for the line tangent to that hyperbola at that given point is y=- \frac{1}{3}x+2y=

Answer 2

The equation of the tangent is x+3y=6

Given:

• Hyperbola $y = \frac{3}{x}$

To find:

• Find the equation of tangent at point (3,1).

Solution:

Concept to be used:

Equation of tangent on curve:

$\bf y - y_1 = \frac{dy}{dx}(x - x_1)$

Step 1:

Find the slope of the tangent.

Slope of the tangent is dy/dx.

Apply power rule of differentiation.

$\frac{dy}{dx} = - \frac{3}{ {x}^{2} }$

put x=3

$\frac{dy}{dx} = \frac{ - 3}{9}$

$\bf m= \frac{ - 1}{3}$

The slope of the tangent line is -1/3.

Step 2:

Find the equation of tangent.

$y - 1 = \frac{ - 1}{3} (x - 3)$

$\implies \: 3(y - 1) = - 1 (x - 3)$

$\implies \: 3y - 3= - x + 3$

$\implies \: x + 3y = 6$

Thus,

The equation of the tangent is x+3y=6.

Learn more :

1) Find the tangent line to the curve y =

√x at x = 4

(1) y=x/4+1

(3) y = 5x + 7

(2) y = 3x -2

(4) y = 4x+8

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2) The point at which the tangent to the curve y= √4x-3 -1 has its slope 2/3. urgent pls

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