Find an Equation of the Tangent Lines Passing Through a Poin

[logan3]

Homework Statement

Find an equation of the tangent lines [in point-slope form] to the hyperbola $x^2 + y^2 = 16$ that pass through the point (2, -2).

Homework Equations

Point-slope form: $y - y_1 = m(x - x_1)$
Slope: $m = \frac {y_1 - y_0} {x_1 - x_0}$

The Attempt at a Solution

First, take the derivative using implicit differentiation.

$D_x: 2x + 2yy' = 0 \Rightarrow y' = \frac {x} {y}$

Since the derivative is the general form of the slope of a line at any of its points, then we can put it equal to the equation for slope. Furthermore, the derivative is tangent to a particular point on the original equation, which we can call point $(x_1, y_1)$. Since we found the equation of the slope at this point to be $y' = \frac {x} {y}$, then plugging in the point $(x_1, y_1)$ (a particular case) should also work: $y' = \frac {x_1} {y_1}$. Finally, we were given our second point above as (2, -2).

$y' = \frac {x_1} {y_1} = \frac {y_1 + 2} {x_1 - 2} \Rightarrow x_1^2 - 2x_1 = y_1^2 + 2y_1 \Rightarrow 2x_1 - 2y_1 = x_1^2 + y_1^2$

Since the general equation $x^2 + y^2 = 16$ works for any pair of points on the line, then it should also work for the particular case of the point on the tangent line that is intersecting it, $(x_1, y_1)$.

$2x_1 - 2y_1 = x_1^2 + y_1^2 = 16 \Rightarrow x_1 - y_1 = 8 \Rightarrow y_1 = x_1 - 8$

Substitute into $x_1^2 + y_1^2 = 16$ and solve for $x_1$ and $y_1$. We find $x_1 = 5$, $y_1 = 3$ and $m = \frac 5 3$. Hence, the equation of the tangent line is $y - 3 = \frac 5 3 (x - 5)$.

It took me awhile to reason through this problem, so I wasn't sure if it was correct. Thank-you.

 

[Mark44]

Homework Statement

Find an equation of the tangent lines [in point-slope form] to the hyperbola $x^2 + y^2 = 16$

What you wrote is a circle. Did you mean $x^2 - y^2 = 16$?

that pass through the point (2, -2).

Homework Equations

Point-slope form: $y - y_1 = m(x - x_1)$
Slope: $m = \frac {y_1 - y_0} {x_1 - x_0}$

The Attempt at a Solution

First, take the derivative using implicit differentiation.

$D_x: 2x + 2yy' = 0 \Rightarrow y' = \frac {x} {y}$

The above isn't consistent with the equation you showed (which itself could be incorrect). If you differentiate $x^2 + y^2 = 16$ w.r.t. x, you get 2x + 2yy' = 0, or y' = -x/y.

Since the derivative is the general form of the slope of a line at any of its points, then we can put it equal to the equation for slope. Furthermore, the derivative is tangent to a particular point on the original equation, which we can call point $(x_1, y_1)$. Since we found the equation of the slope at this point to be $y' = \frac {x} {y}$, then plugging in the point $(x_1, y_1)$ (a particular case) should also work: $y' = \frac {x_1} {y_1}$. Finally, we were given our second point above as (2, -2).

$y' = \frac {x_1} {y_1} = \frac {y_1 + 2} {x_1 - 2} \Rightarrow x_1^2 - 2x_1 = y_1^2 + 2y_1 \Rightarrow 2x_1 - 2y_1 = x_1^2 + y_1^2$

Since the general equation $x^2 + y^2 = 16$ works for any pair of points on the line, then it should also work for the particular case of the point on the tangent line that is intersecting it, $(x_1, y_1)$.

$2x_1 - 2y_1 = x_1^2 + y_1^2 = 16 \Rightarrow x_1 - y_1 = 8 \Rightarrow y_1 = x_1 - 8$

Substitute into $x_1^2 + y_1^2 = 16$ and solve for $x_1$ and $y_1$. We find $x_1 = 5$, $y_1 = 3$ and $m = \frac 5 3$. Hence, the equation of the tangent line is $y - 3 = \frac 5 3 (x - 5)$.

It took me awhile to reason through this problem, so I wasn't sure if it was correct. Thank-you.