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IMG_20190218_081137.jpg 9. The locus of the orthocenter of the triangle formed by the lines (l + p) x (1 + q) x — qy + q (l + q) = O and y = O, where p q, is (a) a hyperbola (b) a parabola (c) an ellipse (d) a straight line (2009)

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given eq of line are

(1 + p)x – py + p(1+p) = 0      ...(i)

(1 + q)x – qy + q(1+q) = 0      ...(ii)

y = 0                                        ...(iii)

solving eq (i) and (ii) we get

C(pq, (1+p)(1+q))

and A(-p,0)  and B(-q,0)

equation of altitude CM passing through C and perpendicular to AB is

 x = p q                                    ..(iv)

slope of line (ii) is (1+q)/q

slope of altitude BN  (N is on line AC) is -q/(1+q)

equation of BN is

                 ....(v)

let H be orthocenter (h,k) intersection of line (iv) and (v)

solve this 

x = pq and y = -pq

so, h=pq and y=-pq

x+y=0

 

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himanshu.meshram

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