/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 80 Two points \(A(1,4)\) and \(B(3,... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 80

Two points \(A(1,4)\) and \(B(3,0)\) are given on the ellipse \(2 x^{2}+y^{2}=18\). The co-ordinates of a point on the ellipse such that the area of the triangle \(A B C\) is greatest is equal to (a) \((\sqrt{6}, \sqrt{6})\) (b) \((-\sqrt{6}, \sqrt{6})\) (c) \((\sqrt{6},-\sqrt{6})\) (d) \((-\sqrt{6},-\sqrt{6})\)

Short Answer

Expert verified
The coordinates of point C that give the maximum area of triangle ABC are \((\sqrt{6}, \sqrt{6})\), which corresponds to option (a).

Step by step solution

01

Equation of the ellipse

We are given the ellipse equation: \[2x^2 + y^2 = 18\] Let the coordinates of point C be \((x_C, y_C)\). Since C lies on this ellipse, \[2x_C^2 + y_C^2 = 18\]
02

Area of triangle ABC

Using the vertices A, B, and C, we can find the area of triangle ABC using the determinant formula: \[Area =\frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|\] Substituting the coordinates of points A (1, 4) and B (3, 0), we get: \[Area = \frac{1}{2} |1(0 - y_C) + 3(y_C - 4) + x_C(4 - 0)| = \frac{1}{2} |-y_C + 3y_C - 12 + 4x_C|\]
03

Simplify the area expression

Simplifying the area expression to get: \[Area = \frac{1}{2} |2y_C - 12 + 4x_C|\]
04

Eliminate y_C using ellipse equation

We want to avoid having two variables in our expression to maximize, so we can solve for y_C and replace it with an expression using only x_C. From the equation of the ellipse, we have: \[y_C^2 = 18 - 2x_C^2\] \[y_C = \sqrt{18 - 2x_C^2}\] Now, replace y_C in the area expression with this new expression: \[Area = \frac{1}{2} |2(\sqrt{18 - 2x_C^2}) - 12 + 4x_C|\]
05

Maximize the area

To maximize the area of triangle ABC, we want to maximize the expression: \[2(\sqrt{18 - 2x_C^2}) - 12 + 4x_C\] Taking the derivative of this expression with respect to x_C and setting it equal to 0 gives us the critical points for which the area is maximized or minimized. The derivative is: \[\frac{d}{dx_C}(2 \sqrt{18 - 2x_C^2} - 12 + 4x_C) = \frac{-4x_C}{\sqrt{18-2x_C^2}} + 4\] Setting the derivative equal to 0 and solving for x_C yields \(x_C = \pm \sqrt{6}\). Substitute the values of x_C back into the equation of the ellipse to find the corresponding y_C values, giving us two points C: \((\sqrt{6}, \sqrt{6})\) and \((- \sqrt{6}, \sqrt{6})\).
06

Determine the point that gives the maximum area

In order to determine which of the two points give the maximum area, we can compare the areas of triangles formed by each pair of points. The points are: 1. (1, 4), (3, 0), and \((\sqrt{6}, \sqrt{6})\) 2. (1, 4), (3, 0), and \((- \sqrt{6}, \sqrt{6})\) Calculating the area of both triangles, we find that the triangle with vertices (1, 4), (3, 0), and \((\sqrt{6}, \sqrt{6})\) has the greatest area. Therefore, the coordinates of point C that give the maximum area of triangle ABC are \((\sqrt{6}, \sqrt{6})\), which corresponds to option (a).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Ellipse Equation
An ellipse is a curve in a plane surrounding two focal points. Its shape is determined by the lengths of the semi-major and semi-minor axes. The standard form of the ellipse equation in a coordinate plane 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. For the exercise, we're working with an equation of the form \[2x^2 + y^2 = 18\] which can be rearranged to fit the standard form. Understanding this equation allows us to find points on the ellipse, as demonstrated by finding the maximum area of a triangle formed by points on the ellipse.
Determinant Formula for Area
The determinant formula provides a way to calculate the area of a triangle given the coordinates of its vertices. The formula for the area of a triangle with vertices at \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by:
\[Area = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\]
When applying this formula in problems like maximizing the area of a triangle with vertices on an ellipse, it simplifies the process significantly as it removes the need for more complex geometric considerations.
Derivative Maximization
Derivative maximization is a technique used in calculus to find the maximum or minimum values of functions. By taking the derivative of a function and equating it to zero, we can find critical points. These points are then analyzed to determine whether they correspond to maximum, minimum, or saddle points. In this exercise, we need to maximize the area of a triangle, so we first express the area as a function of one variable, then take its derivative, and finally set it to zero to solve for the variable.
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, involves the study of geometry using a coordinate system. This allows for algebraic manipulation and the solving of geometric problems. It enables the translation of geometric shapes into algebraic equations, as with the ellipse equation we're using in this exercise. By bringing together algebra and geometry, finding the coordinates of a point on the ellipse that leads to the largest area of a triangle becomes a solvable problem via algebraic methods.
Application of Calculus in Geometry
Calculus can be applied in geometry to solve problems that involve rates of change and accumulation. In the context of maximizing the area of a triangle whose vertices lie on an ellipse, calculus aids in finding the maximum area by providing tools to calculate the derivative and critical points of area functions. Through the application of derivatives, we can find precise solutions to geometric optimization problems, showcasing the power of calculus in the realm of geometric concepts.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The interval in which \(f(x)=\cot ^{-1} x+x\) increases, is (a) \(\mathbb{R}\) (b) \((0, \infty)\) (c) \(\mathbb{R}-\\{n \pi\\}\) (d) None of these

Find all the values of the parameter a for which the points of minima of the function \(f(x)=1+a^{2} x-x^{3}\) satisfy the inequality \(\frac{x^{2}+x+2}{x^{2}+5 x+6}<0\).

The point of the curve \(3 x^{2}-4 y^{2}=72\) which is nearest to the line \(3 x+2 y-1=0\) is: (a) \((-6,3)\) (b) \((6,-3)\) (c) \((6,6)\) (d) \((6,5)\)

If \(f: \mathbb{R} \rightarrow \mathbb{R}, f(x)\) is a differentiable bijective function, then which of the following may be true? (a) \((f(x)-x) f^{\prime \prime}(x)<0 \forall x \in \mathbb{R}\) (b) \((f(x)-x) f^{\prime \prime}(x)>0 \forall x \in \mathbb{R}\) (c) If \((f(x)-x) f^{\prime \prime}(x)>0\), then \(f(x)=f^{-1}(x)\) has no solution (d) If \((f(x)-x) f^{\prime \prime}(x)>0\), then \(f(x)=f^{-1}(x)\) has at least one real solution

Let \(S\) be the set of real values of parameter \(\lambda\) for which the equation \(f(x)=2 x^{3}-3(2+\lambda) x^{2}+12 \lambda x\) has exactly one local maximum and exactly one local minimum. Then \(S\) is a super set of (a) \((-4, \infty)\) (b) \((-3,3)\) (c) \((3, \infty)\) (d) \((-\infty, 0)\)
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.