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
823 views
in Mathematics by (53.1k points)

The ends of a straight line segment (bar)AB of constant length c move on two perpendicular lines OX and OY which are the coordinate axes. If the rectangle OAPB is completed, then show that the locus of the foot of the perpendicular drawn from P on to AB is x2/3 + y2/3 = c2/3 

1 Answer

0 votes
by (53.3k points)
selected by
 
Best answer

See Fig. Let A = (a, 0) and B = (0, b) so that the equation of the line AB is

x/a + y/b = 1   ...(1)

and also 

a2 + b2 = c2   ...(2)

Let M(x1, y1) be the foot of the perpendicular drawn from P onto the line AB. Since M(x1, y1) lies on AB, we have

x1/a + y1/b = 1    ....(3)

Since P = (a, b) and PM is perpendicular to AB, we have

Slope of Slope of PM x AB = -1

Therefore,

(b -y1/a - x1)(b - 0/0 - a) = 1

ax1  - by1 = a2 - b2   .....(4)

From Eqs. (3) and (4), we get

Therefore, the locus of M(x1, y1) is

x2/3 + y2/3 = c2/3

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

...