ABCD is a square of side length a. Its side AB slides between x and y axis in first quadrant. The locus of the foot of perpendicular dropped from the point E on the diagonal AC where E is the midpoint of side AD?

1 Answer
Jan 5, 2018

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#E# is the mid point of side #AD#. EP is the perpendicular on AC.
Let the coordinates of foot of the perpendicular (P) be #(h,k)#

The coordinates of A are #(0,0)#

So length of #AP=sqrt(h^2+k^2)#

In right triangle #APE# ,hypotenuse #AE=a/2# and #/_EAP=45^@#.

So #AP=AE*cos45^@=a/2xx1/sqrt2#

Hence we get #sqrt(h^2+k^2)=a/2xx1/sqrt2#

#=>h^2+k^2=a^2/8#

so locus of foot of the perpendicular will be

#=>x^2+y^2=a^2/8#