Equation of a line is given by #y + 2at = t(x-at^2)#, t being a parameter. Find the locus of the point intersection of the lines which are at right angles?

1 Answer
Mar 16, 2018

# y^2=a(x-3a)#,

Explanation:

Note that, the line #y+2at=t(x-at^2)# has slope #t#.

Consider #2# lines # l_j : y+2at_j=t_j(x-at_j^2), j=1,2#.

Clearly, #l_1nnl_2={(X,Y)}," say"#

#={(a(t_1^2+t_1t_2+t_2^2+2)),at_1t_2(t_1+t_2)}#.

If #l_1botl_2," then, "t_1*t_2=-1," so that, "#

#X=a(t_1^2+(-1)+t_2^2+2)=a(t_1^2+1+t_2^2), and, #

#Y=a(-1)(t_1+t_2)=-a(t_1+t_2)#.

We will use the condition #t_1t_2=-1# to eliminate the parameters

#t_1 and t_2" from "X and Y# to get the desired locus.

We have, #X/a={t_1^2+t_2^2}+1,#

#={(t_1+t_2)^2-2t_1t_2}+1#,

#={(-Y/a)^2-2(-1)}+1#.

#rArr X/a=Y^2/a^2+3," leading to, "y^2=a(x-3a)#,

as the reqd. locus in the conventional #(x,y).#

Feel & Spread the Joy of Maths.!