Question:
The circumcenter of a triangle lies at the origin and its centroid is the midpoint of the line segment joining the points \((a^2+1,a^2+1)\) and \((2a,-2a)\), where \(a\neq0\). Then the equation of the parabola passing through the orthocentre is
The circumcenter of a triangle lies at the origin and its centroid is the midpoint of the line segment joining the points \((a^2+1,a^2+1)\) and \((2a,-2a)\), where \(a\neq0\). Then the equation of the parabola passing through the orthocentre is
Show Hint
A very important triangle relation is:
\[
\vec{OH}=3\vec{OG}.
\]
Whenever circumcentre is at origin, orthocentre can immediately be obtained by tripling centroid coordinates.
Updated On: Oct 6, 2026
- \(y-2ax=0\)
- \(y-(a^2+1)x=0\)
- \(x+y=0\)
- \((a-1)^2x-(a+1)^2y=0\)
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The Correct Option is D
Solution and Explanation
Concept:
For any triangle:
\[
\vec{OH}=3\vec{OG},
\]
where \(O\) is circumcentre, \(G\) is centroid and \(H\) is orthocentre.
Since circumcentre is at origin,
\[
\vec H =3\vec G.
\]
Step 1: Find centroid coordinates.
The midpoint of \[ (a^2+1,a^2+1) \] and \[ (2a,-2a) \] is \[ G\left( \frac{a^2+1+2a}{2}, \frac{a^2+1-2a}{2} \right). \] Simplifying, \[ G\left( \frac{(a+1)^2}{2}, \frac{(a-1)^2}{2} \right). \]
Step 2: Find orthocentre coordinates.
Since \[ H=3G, \] we obtain \[ H\left( \frac{3(a+1)^2}{2}, \frac{3(a-1)^2}{2} \right). \]
Step 3: Find equation through the orthocentre.
The required line passes through the origin and orthocentre. Slope: \[ m= \frac{\frac{3(a-1)^2}{2}} {\frac{3(a+1)^2}{2}} = \frac{(a-1)^2}{(a+1)^2}. \] Hence, \[ y= \frac{(a-1)^2}{(a+1)^2}x. \] Rearranging, \[ (a+1)^2y=(a-1)^2x. \] Therefore, \[ \boxed{(a-1)^2x-(a+1)^2y=0}. \]
Step 1: Find centroid coordinates.
The midpoint of \[ (a^2+1,a^2+1) \] and \[ (2a,-2a) \] is \[ G\left( \frac{a^2+1+2a}{2}, \frac{a^2+1-2a}{2} \right). \] Simplifying, \[ G\left( \frac{(a+1)^2}{2}, \frac{(a-1)^2}{2} \right). \]
Step 2: Find orthocentre coordinates.
Since \[ H=3G, \] we obtain \[ H\left( \frac{3(a+1)^2}{2}, \frac{3(a-1)^2}{2} \right). \]
Step 3: Find equation through the orthocentre.
The required line passes through the origin and orthocentre. Slope: \[ m= \frac{\frac{3(a-1)^2}{2}} {\frac{3(a+1)^2}{2}} = \frac{(a-1)^2}{(a+1)^2}. \] Hence, \[ y= \frac{(a-1)^2}{(a+1)^2}x. \] Rearranging, \[ (a+1)^2y=(a-1)^2x. \] Therefore, \[ \boxed{(a-1)^2x-(a+1)^2y=0}. \]
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