Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 1280

The locus of the variable point whose distance from \((-2,0)\) is \((2 / 3)\) times its distance from the line \(\mathrm{x}=-(9 / 2)\) is (a) ellipse (b) parabola (c) circle (d) hyperbola

Short Answer

Expert verified
The locus of the variable point is an ellipse, which can be derived from the equation \[ \frac{5}{9}x^2 - \frac{4}{3}x + y^2 - 1 = 0 \].

Step by step solution

01

Distance formula

Let the point be \((x, y)\). We need to find the distance between this point and \((-2,0)\), and then between the point and the line \(x=-(9 / 2)\) using the distance formula. The distance between \((x,y)\) and \((-2,0)\) is given by the formula: \[ D_1 = \sqrt{(x + 2)^2 + (y - 0)^2} \] The distance between \((x,y)\) and the line \(x=-(9 / 2)\) is given by the formula: \[ D_2 = |x - (-\frac{9}{2})| = |x + \frac{9}{2}|\]
02

Equate the ratios and simplify

Now we are given that the distance from the point to \((-2,0)\) is \((2 / 3)\) times its distance from the line \(x=-(9 / 2)\). Thus, \[ D_1 = \frac{2}{3}D_2 \] Substitute the values of \(D_1\) and \(D_2\) from step 1. \[ \sqrt{(x + 2)^2 + y^2} = \frac{2}{3}|x + \frac{9}{2}|\] Square both sides to eliminate the square root: \[ (x + 2)^2 + y^2 = \frac{4}{9}(x + \frac{9}{2})^2 \]
03

Expand and simplify

Expand both sides of the equation and simplify to determine the locus. \[ x^2 + 4x + 4 + y^2 = \frac{4}{9}(x^2 + 9x + \frac{81}{4}) \] Multiply both sides of the equation by 9: \[ 9x^2 + 36x + 36 + 9y^2 = 4(x^2 + 9x + \frac{81}{4}) \] Combining terms and simplifying, \[ 5x^2 - 12x + 9y^2 - 9 = 0 \] Divide both sides of the equation by 9: \[ \frac{5}{9}x^2 - \frac{4}{3}x + y^2 - 1 = 0 \] By looking at the equation, we can see that the locus is an ellipse because it fits the form, \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] Thus, the answer is (a) ellipse.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 angle between the lines \(\\{(\mathrm{x}, 0) /(\mathrm{x} \in \mathrm{R})\\}\) and \(\\{(0, y) /(y \in R)\\}\) is \(\ldots \ldots\) (a) \((\pi / 2)\) (b) \(-(\pi / 2)\) (c) 0 (d) \(\pi\)

If \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are in A.P. then \(\mathrm{ax}+\mathrm{by}+\mathrm{c}=0\) represents (a) a single line (b) a family of concurrent lines (c) a family of parallel lines (d) a family of circle

If the lengths of perpendicular drawn from the origin to the lines \(x \cos \alpha-y \sin \alpha=\sin 2 a \alpha\) and \(x \sin \alpha+y \cos \alpha=\cos 2 \alpha\) are \(p\) and \(q\) respectively, then \(p^{2}+q^{2}=\ldots \ldots\) (a) 4 (b) 3 (c) 2 (d) 1

\(\mathrm{A}(4,0), \mathrm{B}(0,3), \mathrm{C}(6,1)\) be vertices of triangle \(\mathrm{ABC}\). Slope of bisector of angle \(\mathrm{C}\) will be (a) \(3 \sqrt{2}-7\) (b) \(5 \sqrt{2}-7\) (c) \(6 \sqrt{2}-7\) (d) none

If \(\mathrm{P}(-1,0), \mathrm{Q}(0,0)\) and \(\mathrm{R}(3,3 \sqrt{3})\), then the equation of bisector of \(\angle P Q R\) is \(\ldots \ldots .\) (a) \((\sqrt{3} / 2) \mathrm{x}+\mathrm{y}=0\) (b) \(x+(\sqrt{3} / 2) y=0\) (c) \(\sqrt{3 x+y}=0\) (d) \(x+\sqrt{3} y=0\)
See all solutions

Recommended explanations on English Textbooks

Free Response Essay

Read Explanation

Phonetics

Read Explanation

Research and Composition

Read Explanation

Language Acquisition

Read Explanation

Textual Analysis

Read Explanation

English Language Study

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.

Sign-up for free