Question

If PQ be a normal chord of the parabola and if S be the focus, prove that the locus of the centroid of the triangle SPQ is curve $36a{y}^2\left(3x-5a\right)-81y^4=128a^4$.

Answer 1

Hint- Always remember the equation of the normal chord. And put the values as $t_1,t_2$. We will solve the question by using the formula of the centroid of a triangle SPQ given the coordinates are $\left(h,k\right)$. The formula will be $h={\displaystyle \frac{a+a{t}_1^2+a{\left(t_1+{{\displaystyle \frac{2}{t}}}_1\right)}^2}{3}}$, $k={\displaystyle \frac{0+2a{t}_1-2a{t}_1-{\displaystyle \frac{4a}{t_1}}}{3}}$

Complete Step-by-step answer:

If the focus of the parabola is S, then the coordinates will be $\left(a,0\right)$.
As we all know, the equation of a parabola is $y^2=4ax$.
Let the equation of the normal chord at point P be $P\left(a{t}_1^2,2a{t}_1\right)$.
Then, the value of y will be-
$y=t_{1}x-2a{t}_1-a{t}_1^3$
Let the other point where it cuts be named Q, then the coordinates of the point Q will be $Q\left(a{t}_2^2,2a{t}_2\right)$.
The relation here is-
$t_2=-t_1-{\displaystyle \frac{2}{t_1}}$
The formula for the centroid of the triangle SPQ where the coordinates are $\left(h,k\right)$is:
 $h={\displaystyle \frac{a+a{t}_1^2+a{\left(t_1+{{\displaystyle \frac{2}{t}}}_1\right)}^2}{3}}$
$k={\displaystyle \frac{0+2a{t}_1-2a{t}_1-{\displaystyle \frac{4a}{t_1}}}{3}}$
$\Rightarrow t_1={\displaystyle \frac{-4a}{3k}}$
Now, as we know,
$h={\displaystyle \frac{3a+2a{t}_1^2+{\displaystyle \frac{4a}{t_1^2}}}{3}}$
So,
$3h=3a+2a{\left({\displaystyle \frac{-4a}{3k}}\right)}^2+{\displaystyle \frac{4a}{{\left({\displaystyle \frac{-4a}{3k}}\right)}^2}}$
Replacing $\left(h,k\right)$ by $\left(x,y\right)$, we get this:
$36a{y}^2\left(3x-5a\right)-81y^4=128a^4$
Hence proved.

Note: Use the equation of parabola in the starting of the question and pay special attention to the superscripts and subscripts as they are a wee but congested in these type of questions and mat completely vary your answer if not done in a right way.