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Find equations of both tangent lines to the graph of the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\) that pass through the point \((4,0)\) not on the graph.

Short Answer

Expert verified
The equations of the tangent lines to the graph of the ellipse that pass through the point (4,0) are \(y=\frac{1}{4}(x-4)\) and \(y=-\frac{1}{4}(x-4)\).

Step by step solution

01

Equation of line

First, let's find the equation of a line passing through the given point \((4,0)\). The equation of such a line is \(y-0=m(x-4)\) where \(m\) is the slope of the line.
02

Substituting into the ellipse equation

Second, we substitute this equation into the equation of the ellipse. This will give us \(\frac{x^{2}}{4}+\frac{[(m(x-4))^{2}]}{9}=1\). Simplify this to obtain the quadratic equation in x, \((1+4m^{2})x^{2}-(16m^{2})x+16m^{2}-4=0\). The general form of a second degree equation is \(ax^{2}+bx+c=0\).
03

Solving quadratic equation

Third, we solve the quadratic equation for \(m\). Since the lines are tangents, there are equal roots to the equation. For equal roots, the discriminant of the quadratic equation (\(b^{2}-4ac\)) must be equal to zero. We set \((16m^{2})^{2}-4(1+4m^{2})(16m^{2}-4)=0\) and solve for \(m\) to get \(m=\frac{1}{4}\) and \(m=-\frac{1}{4}\).
04

Finding the equations of the tangents

Fourth, we substitute these values of \(m\) into the equation of the line \(y-0=m(x-4)\) to find the equations of the tangents. Substituting \(m=\frac{1}{4}\) gives \(y=\frac{1}{4}(x-4)\) and substituting \(m=-\frac{1}{4}\) gives \(y=-\frac{1}{4}(x-4)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ellipse Equation
When discussing curves in coordinate geometry, an ellipse is a fundamental figure. It is the set of all points where the sum of the distances from two fixed points, called foci, is constant. The general form of an ellipse equation is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively.

An ellipse equation allows us to describe the boundary of an ellipse and discern the shape, orientation, and size. Solving problems involving tangents to an ellipse often involves inserting the linear equation of a tangent line into the quadratic format of the ellipse equation and then finding specific conditions for the line to only touch the ellipse at one point, reflecting properties of tangent lines.
Quadratic Equation
A quadratic equation is a second-order polynomial equation in a single variable \(x\) with a coefficient of \(x^2\) that is not zero. It can be written in the standard form as \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a \eq 0\).

The solutions to a quadratic equation are given by the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\), which depends crucially on the value of the discriminant, \(b^2-4ac\). These solutions represent the \(x\)-intercepts of the parabola represented by the quadratic equation. Understanding the nature of these solutions is crucial when dealing with problems that require finding the conditions under which a line will tangent an ellipse, such as in the original exercise.
Discriminant of a Quadratic Equation
The discriminant in a quadratic equation is the expression under the square root in the quadratic formula: \(b^2 - 4ac\). It provides critical information about the nature and number of solutions. For instance:
  • If the discriminant is positive, the equation has two real and distinct solutions.
  • If the discriminant is zero, the equation has exactly one real solution, or in geometric terms, the parabola touches the \(x\)-axis just once. This condition is useful when determining the slope of the tangent to an ellipse.
  • If the discriminant is negative, there are no real solutions, but two complex solutions.

In context with tangents, if we are given the task to determine the slope of a line tangent to an ellipse, we look for conditions where the discriminant of the derived quadratic equation is zero. This reflects the fact that the tangent line meets the ellipse at exactly one point.
Slope of a Line
The slope of a line is a measure of its steepness, often represented by the letter \(m\). It is calculated as the ratio of the rise (vertical change) to the run (horizontal change) between two points on the line. The slope is defined by the formula \(m = \frac{\Delta y}{\Delta x}\) where \(\Delta y\) and \(\Delta x\) represent the differences in the \(y\)- and \(x\)-coordinates of two points.

Finding the slope is a crucial part of the process of determining the equation of a line. In the original exercise, the slope of the tangent line to the ellipse is found by setting the discriminant of the quadratic equation, derived by substituting the line equation into the ellipse equation, to zero. The resulting slope tells us how the tangent line is positioned in relation to the ellipse.

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Most popular questions from this chapter

Linear and Quadratic Approximations The linear and quadratic approximations of a function \(f\) at \(x=a\) are $$\begin{array}{l}{P_{1}(x)=f^{\prime}(a)(x-a)+f(a) \text { and }} \\\ {P_{2}(x)=\frac{1}{2} f^{\prime \prime}(a)(x-a)^{2}+f^{\prime}(a)(x-a)+f(a)}\end{array}$$ In Exercises 123 and \(124,\) (a) find the specified linear and quadratic approximations of \(f,(b)\) use a graphing utility to graph \(f\) and the approximations, (c) determine whether \(P_{1}\) or \(P_{2}\) is the better approximation, and (d) state how the accuracy changes as you move farther from \(x=a\) . $$ f(x)=\tan x ; \quad a=\frac{\pi}{4} $$

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