Guest

Show that the locus of a point which moves in such a way that the square of its distance from the base of an isosceles triangle is equal to the rectangle under its distance from other sides, is a circle.

Show that the locus of a point which moves in such a way that the square of its distance from the base of an isosceles triangle is equal to the rectangle under its distance from other sides, is a circle.

Grade:11

1 Answers

AJIT AskiitiansExpert-IITD
68 Points
13 years ago

Dear Aman ,

I can do this by coordinate geometry, but I don't see a synthetic (euclidean) proof of it. If the triangle is PQR, with PQ = PR, take P, Q, R to be the points (0,b), (–a,0), (a,0) respectively. The equation of the line PQ is and the equation of PR is . If I understand the question correctly, you want to find the locus of a point X = (x,y) which moves so that the square of its distance from QR is equal to the product of its distances from PQ and PR. With the above choices of P, Q and R, the distance from X to PQ is , and the distance from X to PR is . The distance from X to QR is |y|. So the locus of X is given by the equation . This simplifies to , which represents a circle with centre at and radius . In terms of the geometry of the picture, this circle has its centre C on the perpendicular bisector of QR. It passes through Q and R, and is tangential to PQ and PR. 

 

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively : Click here to download the toolbar..

Askiitians Expert

Ajit Singh Verma IITD

Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free