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?

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Answer 1 · 2017-12-15T20:04:46

#x = 1/a y^2+3a#

#L_1 = y-tx+2at+at^3 = 0#

and making #t -> -1/t#

#L_2 = y+x/t-(2a)/t-a/t^3 = 0#

Clearly #L_1 bot L_2#

Solving #L_1,L_2# for #x,y# we obtain

#{(x = a(t^2+1/t^2+1) = a((1/t-t)^2+3)),(y = a(1/t-t)):}#

and now making #xi = 1/t-t# we have

#{(x = a(xi^2+3)),(y=a xi):}#

and after eliminating #xi#

we obtain the parabola

#x = 1/a y^2+3a#