Introduction to Ellipse and Tangent Lines

The ellipse is a fundamental conic section characterized by its oval shape. In mathematics, the equation of an ellipse is often given in the form $frac{x^2}{a^2} frac{y^2}{b^2} 1$. However, as we will explore in this article, it is also insightful to explore the tangent and normal lines to an ellipse using implicit differentiation. This method allows us to find the slopes of these lines at specific points on the ellipse.

Verifying the Point on the Ellipse

To find the equations of the tangent and normal lines at a specific point on the ellipse, we must first confirm if the point lies on the ellipse. This involves substituting the coordinates of the point into the ellipse's equation and ensuring that the equality holds.

Step 1: Verify the Point vs. Example Point 3-2

Let's consider the ellipse defined by the equation $4x^2 9y^2 35$. We aim to find the tangent and normal lines at the point (3, -2).

Substituting (3, -2) into the equation, we get:

$4(3)^2 9(-2)^2 4(9) 9(4) 36 36 72$

Since 72 ≠ 35, the point (3, -2) does not lie on the ellipse.

Step 2: Finding a Valid Point on the Ellipse

If we instead take the ellipse $4x^2 9y^2 72$, we can solve for y when x 3 to find a valid point on the ellipse. Rearranging the equation:

$4(3)^2 9y^2 72$

$36 9y^2 72$

$9y^2 36$

$y^2 4$

$y ±2$

This gives us the points (3, 2) and (3, -2). For our purposes, we can use (3, -2) as a valid point on the ellipse.

Finding the Slope of the Tangent Line

We can find the slope of the tangent line by implicitly differentiating the equation of the ellipse. The equation of the ellipse is:

$4x^2 9y^2 72$

Using implicit differentiation, we get:

$8x 18y frac{dy}{dx} 0$

$18y frac{dy}{dx} -8x$

$frac{dy}{dx} -frac{4x}{9y}$

Computing the Slope at Point (3, -2)

Plugging in the point (3, -2) gives us:

$frac{dy}{dx} -frac{4(3)}{9(-2)} frac{12}{18} frac{2}{3}$

Therefore, the slope of the tangent line is $frac{2}{3}$.

Equation of the Tangent and Normal Lines

With the slope of the tangent line known, we can use the point-slope form to write the equations of the tangent and normal lines.

Equation of the Tangent Line

The equation of the tangent line is:

$y - (-2) frac{2}{3}(x - 3)$

$y 2 frac{2}{3}x - 2$

$y frac{2}{3}x - 4$

Equation of the Normal Line

The normal line is perpendicular to the tangent line. The slope of the normal line is the negative reciprocal of the tangent line's slope.

$m_{text{normal}} -frac{3}{2}$

The equation of the normal line is:

$y - (-2) -frac{3}{2}(x - 3)$

$y 2 -frac{3}{2}x frac{9}{2}$

$y -frac{3}{2}x frac{5}{2}$

Conclusion

In conclusion, we have shown how to find the equations of the tangent and normal lines at a specific point on an ellipse using implicit differentiation. This method is not only useful for solving problems in geometry but also provides a deeper understanding of the properties of ellipses. If you have further questions or wish to explore more related concepts, please feel free to ask!