Given the equation of the line :

x^2 + y^2 = 34

We need to find the tangent line at the point ( -3,5)

First we will differentiate with respect to x.

==> 2x + 2yy' = 0

==> 2yy' = -2x

==> y' = -2x/2y

==> y' = -x/y

...

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Given the equation of the line :

x^2 + y^2 = 34

We need to find the tangent line at the point ( -3,5)

First we will differentiate with respect to x.

==> 2x + 2yy' = 0

==> 2yy' = -2x

==> y' = -2x/2y

==> y' = -x/y

Now we will substitute with the point (-3,5) to find the slope.

==> m = y' = -(-3)/5 = 3/5

Then the equation of the line is given by :

y-y1 = m(x-x1)

==> y-5 = (3/5) (x+3)

==> y= (3/5)x + 9/5 + 5

==> y= (3/5)x + 34/5

Then the equation of the line is:

**==> 5y - 3x -34 = 0 **

The slope of the tangent to a curve f(x) is the value of the first derivative of the curve for the appropriate x.

Here we have x^2 + y^2 = 34

Differentiating both the sides 2x + 2y (dy/dx) = 0

=> dy/dx = -2x/2y = -x/y

At x = -3 and y = 5

dy/dx = 3/5

The equation of the tangent is (y - 5) / (x + 3) = 3/5

=> 5y - 25 = 3x + 9

=> 3x - 5y + 34 = 0

**The required equation of the tangent is 3x - 5y + 34 = 0**