JEE Mathematics 2023 Question with Solution

Given: The parabola is $y^2 = 20x$, the line is $y = mx + c$, the focus is $R$, and the centroid of triangle $PQR$ is $G(10,10)$. Also, $c-m=6$.

Find: The value of $(PQ)^2$.

For the parabola $y^2 = 20x$, we compare with $y^2 = 4ax$, so $4a=20$ and hence $a=5$. Therefore, the focus is

Let the intersection points be $P(x_1,y_1)$ and $Q(x_2,y_2)$. Using the centroid formula for triangle $PQR$,

Thus,

Now use the line equation in the form

Substitute this into the parabola $y^2=20x$:

So,

Dividing by $m$,

Since the roots of this quadratic in $y$ are $y_1$ and $y_2$, by Vieta's relations,

But from the centroid condition, $y_1+y_2=30$. Hence,

Given $c-m=6$, we get

Substitute these values into

Solving,

so

Now find the corresponding $x$-coordinates from $x=\frac{y^2}{20}$:

Hence the points are

Therefore,

Therefore, the value of $(PQ)^2$ is $325$. The correct option is A.