Question

Find the equation of the tangent at the point (a,b) to the curve xy = c^2 .

Answer 1

xy = c²
differentiating the above curve at (a,b) we get slope of that tangent at that point
x*dy/dx + y = 0
x*dy/dx = -y
 dy/dx= -y/x
  slope at the point (a,b)
 dy/dx = -b/a
the equation of the tangent at (a,b)
(y-b) = -b/a (x-a)
a(y-b)= -b(x-a)
ay-ab = -bx +ab
bx+ay -2ab = 0 
This is equation of tangent at the point (a,b)

Answer 2

Answer:

The equation of the tangent of the curve xy = $c^{2}$ at the point (a,b) is $\frac{x}{a} + \frac{y}{b} = 2$.

Tangent of a curve: At a given point on a curve, the gradient of the curve is equal to the gradient of the tangent to the curve.

Step-by-step explanation:

Given, the equation of the curve is xy = $c^{2}$.

To obtain the tangent of the curve. compute the derivative with respect to y.

Here we will obtain the following equation,

                                     $x\frac{dx}{dy} + y = 0$

Solving the above equation we obtain $\frac{dy}{dx} = -y/x$.

The slope of the curve at the point (a,b) is -b/a.

Now the equation of the tangent at the point (a,b) is

                            $(y-b) = \frac{-b}{a} (x - a)$

simplifying the above equation

we obtain the equation as $\frac{x}{a} + \frac{y}{b} = 2$.