The locus of the orthocentre of the triangle formed by the lines
( 1 + p )x - py + p( 1 + p ) = 0,
( 1 + q )x - qy + q( 1 + q ) = 0,
and y=0 (where p not equal to q ) is
A) Hyperbola B) parabola C) ellipse D) straight line
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2 Answers
62(1+1/p)x-y+(1+p)=0
(1+1/q)x-y+(1+q)=0
Substracting,
(1/p-1/q)x+(p-q)=0
x=pq
y=(1+p)(1+q)
At the orthocenter, x=pq, becasue we are dropping a perpendicular to the line y=0 from the ponit (pq, (1+p)(1+q))
Also, the other point of intersection would be x=-p, y=0 and (-q,0)
now the perpendicular would be...
px+(1+p)y+pq=0
which has x=pq
so y= -pq
thus the orthocenter is (pq, -pq) which is a straight line...
- Shouvik Mukherjee How the equation px+(1+p)y+pq=0 comes?????Upvote·0· Reply ·2017-02-18 22:07:32
62somehow more than just this meets my eye.. but am not able to use it
The equation fo the two lines is of the form
x/p-y/(1+p)+1=0
whcih is the parametric form..
