If a triangle is formed by any three tangents of the parabola y^2=4ax , two of whose vertices lie on the paranoia x^2=4by , then find the locus of the third vertex . Please do the question by using two methods . I have seen the solution but not satisfied yet so i want another methods to solve it. Share with your friends Share 25 Neha Sethi answered this Dear student Using the parametrization at2,2at, let the three points on the parabola y2=4ax is beap2,2ap,aq2,2aq,ar2,2arFirst find the equation of tangent at ap2,2apThe slope of tangent to y2=4ax at x,y is,2ydydx=4adydx=2ayAt the point ap2,2ap, the slope of tangent is ,dydx=2a2ap=1pSo, equation of tangent at ap2,2ap is given by,y-2ap=1px-ap2py-2ap2=x-ap2py=x-ap2+2ap2py=x+ap2In a similar way, the equation of other two tangents are,qy=x+aq2ry=x+ar2Find the intersection points of these tangents to get,apqq,ap+q,apq,ap+r,aqr,aq+rSince the first two points lie on the parabola x2=4by, we geta2p2q2=4abp+q ....(1)a2p2r2=4abp+r ....(2)The intersection of the remaining pair is given byx=aqr ...(3)y=aq+r ...(4)The locus of the third vertex is found by eliminating p,q,r from (1)-(4)Apply (1)-(2) to get,a2p2q2-r2=4abq-r ⇒a2p2q-rq+r=4ab(q-r)⇒p2=4abq-ra2q-rq+r⇒p2=4baq+r⇒p2=4by [using 4]Apply (1)÷(2) to get,q2r2=p+qp+r⇒q2p+r=r2p+q⇒q2p+q2r=r2p+r2q⇒pq2-r2=-qrq-r⇒p=-qrq+r⇒p=-xy [using 3 and 4]This further gives,x2y2=4by⇒x2=4bySo, the third vertex lies on the same parabola. Regards 38 View Full Answer