ISI MStat 2016 PSA Problem 9 | Equation of a circle

This is a problem from ISI MStat 2016 PSA Problem 9 based on equation of a circle. First, try the problem yourself, then go through the sequential hints we provide.

Equation of a circle- ISI MStat Year 2016 PSA Question 9

Given $\theta$ in the range $0 \leq \theta<\pi,$ the equation $2 x^{2}+2 y^{2}+4 x \cos \theta+8 y \sin \theta+5=0$ represents a circle for all $\theta$ in the interval

$0 < \theta <\frac{\pi}{3}$
$\frac{\pi}{4} < \theta <\frac{3\pi}{4}$
$0 < \theta <\frac{\pi}{2}$
$0 \le \theta <\frac{\pi}{2}$

Key Concepts

Equation of a circle

Trigonometry

Basic Inequality

Check the Answer

Answer: is $\frac{\pi}{4} < \theta <\frac{3\pi}{4}$

ISI MStat 2016 PSA Problem 9

Precollege Mathematics

Try with Hints

Complete the Square.

We get ,

$2{(x+\cos \theta)}^2 + 2{(y+ 2\sin \theta)}^2 = (6{\sin \theta}^2-3))$
$6{\sin \theta}^2-3 > 0 \Rightarrow {\sin^2 \theta} \geq \frac{1}{2}$

We are given that $0 \leq \theta<\pi,$ . So, ${\sin^2 \theta} \geq \frac{1}{2}$ $\Rightarrow \frac{\pi}{4} < \theta <\frac{3\pi}{4}$ .

Similar Problems and Solutions

ISI MStat 2016 PSA Problem 9 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Related

This is a problem from ISI MStat 2016 PSA Problem 9 based on equation of a circle. First, try the problem yourself, then go through the sequential hints we provide.

Equation of a circle- ISI MStat Year 2016 PSA Question 9

Given $\theta$ in the range $0 \leq \theta<\pi,$ the equation $2 x^{2}+2 y^{2}+4 x \cos \theta+8 y \sin \theta+5=0$ represents a circle for all $\theta$ in the interval

$0 < \theta <\frac{\pi}{3}$
$\frac{\pi}{4} < \theta <\frac{3\pi}{4}$
$0 < \theta <\frac{\pi}{2}$
$0 \le \theta <\frac{\pi}{2}$

Key Concepts

Equation of a circle

Trigonometry

Basic Inequality

Check the Answer

Answer: is $\frac{\pi}{4} < \theta <\frac{3\pi}{4}$

ISI MStat 2016 PSA Problem 9

Precollege Mathematics

Try with Hints

Complete the Square.

We get ,

$2{(x+\cos \theta)}^2 + 2{(y+ 2\sin \theta)}^2 = (6{\sin \theta}^2-3))$
$6{\sin \theta}^2-3 > 0 \Rightarrow {\sin^2 \theta} \geq \frac{1}{2}$

We are given that $0 \leq \theta<\pi,$ . So, ${\sin^2 \theta} \geq \frac{1}{2}$ $\Rightarrow \frac{\pi}{4} < \theta <\frac{3\pi}{4}$ .

Similar Problems and Solutions

ISI MStat 2016 PSA Problem 9 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Related

Leave a ReplyCancel reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy

Cheenta Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

HALL OF FAME BLOG OFFLINE CLASS

CHEENTA ON DEMAND BOSE OLYMPIAD

Cheenta Academy Private Limited |

support@cheenta.com

Menu