Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
2.4k views
in Mathematics by (93.6k points)
closed by
Find the area of the figure enclosed by the curve `5x^2+6x y+2y^2+7x+6y+6=0.`

1 Answer

0 votes
by (94.6k points)
selected by
 
Best answer
Equation of curve can be re-written as
`2y^(2)+6(1+x)y+5x^(2)+7x+6=0`
`rArr" "y_(1)=(-3(1+x)-sqrt((3-x)(x-1)))/(2)`
`y_(2)=(-(1+x)+sqrt((3-x)(x-1)))/(2)`
Therefore, the curves `(y_(1) and y_(2))` are defined for values of x for whih `(3-x)(x-1)ge 0,` i.e., `1le x le 3.`
(Actually the given equation represents an ellipse, because `Delta ne 0 and h^(2) lt ab.`)
image
Required are will be given by
`A=overset(3)underset(1)int |y_(1)-y_(2)|dxrArrA=overset(3)underset(1)intsqrt((3-x)(x-1))dx`
`"Put "x=3 cos^(2)theta+ sin^(2)theta, i.e., dx=-2 sin 2theta d""theta`
`therefore" "A=2overset(pi//2)underset(0)intsin^(2) 2theta d""theta =(pi)/(2)` sq. units.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...