Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

Through the point Pα,β, where αβ>0, the straight line xa+yb=1 is drawn so as to for a triangle of area S with the axes. If ab>0, then the least value of S is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

a

αβ

b

2αβ

c

3αβ

d

none

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

OA=OQ+QA=α+βcotθOB=OR+RB=β+αtanθNow, the area of ΔOAB is 12α+βcotθβ+αtanθ=122αβ+α2tanθ+β2cotθ=122αβ+αtanθ−βcotθ+2αβ≥122αβ+2αβ=2αβLeast value of S=2αβ
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon
personalised 1:1 online tutoring