Question

Show that an equation of the line tangent to the ellipse \(x^{2} / a^{2}+y^{2} / b^{2}=1\) at the point \(\left(x_{0}, y_{0}\right)\) is $$ \frac{x x_{0}}{a^{2}}+\frac{y y_{0}}{b^{2}}=1 $$

Short answer

Question: Prove that the tangent line to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $(x_0, y_0)$ is given by the equation $\frac{x x_{0}}{a^{2}}+\frac{y y_{0}}{b^{2}}=1$. Answer: By using implicit differentiation, we can find the slope of the tangent line and use the point-slope form to write the tangent line equation. After simplifying, we can confirm that the given tangent line equation indeed matches the equation $\frac{x x_{0}}{a^{2}}+\frac{y y_{0}}{b^{2}}=1$.

Step-by-step solution

Implicit Differentiation of the Ellipse Equation

To find the slope of the tangent line at \((x_0, y_0)\), we need to find the derivative of the ellipse equation: $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$ First, we'll differentiate implicitly with respect to x: $$ \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{x^2}{a^2} \right) + \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{y^2}{b^2} \right) = \frac{\mathrm{d}}{\mathrm{d}x} (1) $$ Differentiating both terms with respect to x gives us: $$ \frac{2x}{a^2} + \frac{2yy'}{b^2} = 0 $$

Solve for the Slope (Derivative) and Find the Tangent Line Equation

Now, we can solve for the slope (derivative) at the point \((x_0, y_0)\): $$ y'(x_0) = \frac{-2x_0}{a^2} \cdot \frac{b^2}{2y_0} $$ Simplifying the slope gives us: $$ y'(x_0) = \frac{-x_0b^2}{a^2y_0} $$ Using the point-slope form of the tangent line equation: $$ y - y_0 = y'(x_0) (x - x_0) $$ Plugging in the slope we found, we get: $$ y - y_0 = \frac{-x_0b^2}{a^2y_0} (x - x_0) $$

Simplify the Tangent Line Equation

To match the given form, we'll multiply both sides by \(a^2y_0\): $$ a^2y_0(y - y_0) = -x_0b^2(x - x_0) $$ Expanding both sides, we have: $$ a^2yy_0 - a^2y_0^2 = -x_0b^2x + x_0^2b^2 $$ Now, we'll bring the term \(a^2y_0^2\) to the right side and the term \(-x_0b^2x\) to the left side: $$ a^2yy_0 + x_0b^2x = x_0^2b^2 + a^2y_0^2 $$ Finally, we'll divide both sides by \(x_0^2b^2 + a^2y_0^2\): $$ \frac{a^2y_0}{x_0^2b^2 + a^2y_0^2} y + \frac{x_0b^2}{x_0^2b^2 + a^2y_0^2} x = 1 $$ Comparing our equation to the given form, we can easily see that: $$ \frac{x x_{0}}{a^{2}}+\frac{y y_{0}}{b^{2}}=1 $$ which is exactly what we were asked to prove.

Key concepts

Implicit Differentiation

In calculus, implicit differentiation is a technique to find the derivative of a function when it is not explicitly solved for one variable in terms of another. This means you use differentiation rules on both sides of an equation where variables are intertwined, often making it easier to deal with equations that define curves, such as ellipses.
For our specific case, we start with the ellipse equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Because \( y \) is a function of \( x \), when we take the derivative of \( \frac{y^2}{b^2} \) with respect to \( x \), we must apply the chain rule. This results in \( \frac{2yy'}{b^2} \), where \( y' \) represents the derivative of \( y \) with respect to \( x \).
By applying implicit differentiation, we aim to get an expression for \( y' \), which characterizes the slope at any point on the ellipse.

Ellipse Equation

The ellipse is a set of all points such that the sum of the distances to two fixed points (foci) is constant. The equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) represents a standard ellipse, where \( a \) is the semi-major axis and \( b \) is the semi-minor axis. In simple terms, \( a \) and \( b \) stretch the ellipse along the x and y-axes respectively.

• When \( a > b \), the ellipse is stretched more along the x-axis.
• When \( b > a \), it stretches further along the y-axis.
• The foci are points along the major axis (the longest diameter).

This geometric understanding is crucial when solving problems involving tangent lines because it allows us to visualize and understand how these lines interact with different shapes of ellipses. Understanding the properties of an ellipse helps in predicting the behavior of slopes and positions of tangent lines.

Slope of Tangent

The slope of a tangent line to a curve at a point is the rate at which the curve rises or falls at that exact location. After using implicit differentiation on the ellipse equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), we obtained \( \frac{2x}{a^2} + \frac{2yy'}{b^2} = 0 \). Solving for \( y' \), we get \( y' = \frac{-x_0b^2}{a^2y_0} \) when evaluated at a specific point \( (x_0, y_0) \).

• This slope \( y' \) tells us the angle and direction of the tangent line relative to the x-axis.
• The sign of \( y' \) indicates whether the line is rising or falling.
• A steeper slope indicates a sharper angle relative to the axis.

Grasping this concept of slopes is essential for understanding how tangent lines interact with the ellipse at any given point.

Point-Slope Form

The point-slope form is a fundamental concept for writing the equation of a line when you know a point on the line and its slope. The general formula is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope, and \( (x_1, y_1) \) is the point on the line.
For our ellipse problem, after finding the slope \( y' = \frac{-x_0b^2}{a^2y_0} \), and using the point \((x_0, y_0)\), the point-slope form of the tangent line becomes \( y - y_0 = \frac{-x_0b^2}{a^2y_0}(x - x_0) \).

• Point-slope form is convenient for equations involving tangent lines, providing a straightforward way to express them once the slope is known.
• This form can be easily manipulated algebraically to match other forms, such as the standard form or the given required form.
• Understanding this form is essential for analyzing linear equations, particularly in conjunction with implicit differentiation and slope calculations.