A) a hyperbola
B) a parabola
C) an ellipse
D) a straight line
Correct Answer: D
Solution :
We have, | |
\[\left( 1+p \right)x-py+p\left( 1+p \right)=0\] | ??. (i) |
\[\left( 1+q \right)x-qy+q\left( 1+q \right)=0\] | ?? (ii) |
\[y=0\] | ..?. (iii) |
From Eqs. (i) and (ii), we get | |
\[x=pq,y=\left( p+1 \right)\left( q+1 \right)\] | |
From Eqs. (i) and (iii), we get | |
\[x=-p,y=0\] | |
From Eqs. (ii) and (iii), we get | |
\[x=-q,y=0\] |
Equation of altitude from C to AB is |
\[x=pq\] |
Altitude from B to AC is |
\[y=\frac{-q}{1+q}\left( x+p \right)\] |
Put \[x=pq\], we get \[y=-pq\] |
\[\because \,\,\,\,h=pq,k=-pq\] |
\[\therefore \,\,h+k=0\] |
Locus of orthocenter \[x=y=0\] which represent the equation of straight line. |
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