Question

If vertices of $△ABC$ are $A(3,4)B(5\sin t,5cost)$ and $C(5\cos t,-5\sin t)$, then find the locus of orthocentre of $△ABC$.

Answer 1

The orthocenter is the intersection of altitudes of a $△ABC$

Now, let the point $O(0,0)$ equidistance from $A,B$ and $C$

$OA=\sqrt{(3-0{)}^2+(4-0{)}^2=\sqrt{9+16}}=\sqrt{25(}=5$

$OB=\sqrt{(5\sin t-0{)}^2+(5\cos t-0{)}^2}=\sqrt{25{\cos }^{2}t+25{\sin }^{2}t}=\sqrt{25\left(1\right)}=5$

$OC=\sqrt{(5\cos t-0{)}^2+(5\cos \sin t-0{)}^2}=\sqrt{25{\cos }^{2}t+25{\sin }^{2}t}=\sqrt{25\left(1\right)}=5$

So, we can say $O(0,0)$ is circumcentre of $△ABC$

Now, let $H(h,k)$ be the orthocentre of $△ABC,△G$ is centroid

$G={(\frac{3+5\sin t+5\cos t}{3},\frac{4+5\cos t-5\sin t}{3})}_{.}...\left(1\right)$

$G$ divides $H$ and $O$ in ratio $2:1$

Coordinate of $G=(\frac{h}{3},\frac{k}{3})....\left(2\right)$

Comparing (1) and (2), we get

$h=3+5\sin t+5\cos t$ and $k=4+5\cos t-5\sin t$

$\Rightarrow (h-3)=5\sin t+5\cos t$ and $(k-4)=5\cos t-5\sin t$.