JEE Mathematics 2025 Question with Solution

Given: The circle is $x^2 + y^2 - 8x = 0$, the hyperbola is $\frac{x^2}{9} - \frac{y^2}{4} = 1$, and point $P(x_1,y_1)$ lies on the line $2x_1 - 3y_1 + 4 = 0$.

Find: The line on which the centroid of $\triangle PAB$ lies, where $A$ and $B$ are the intersection points of the circle and hyperbola.

Rewrite the circle as

$$(x-4)^2 + y^2 = 16$$

so

$$y^2 = 8x - x^2.$$

Substitute this into the hyperbola:

$$\frac{x^2}{9} - \frac{8x - x^2}{4} = 1$$ $$\frac{x^2}{9} - 2x + \frac{x^2}{4} = 1$$

Taking LCM $36$,

$$4x^2 - 72x + 9x^2 = 36$$ $$13x^2 - 72x - 36 = 0$$

Solving,

$$x = \frac{72 \pm \sqrt{(-72)^2 - 4(13)(-36)}}{26} = \frac{72 \pm \sqrt{7056}}{26} = \frac{72 \pm 84}{26}$$

Hence,

$$x = 6 \quad \text{or} \quad x = -\frac{6}{13}.$$

Using $x=6$ in $y^2 = 8x - x^2$,

$$y^2 = 48 - 36 = 12$$

so the required intersection points are

$$A(6, 2\sqrt{3}), \quad B(6, -2\sqrt{3}).$$

Now the centroid $G$ of $\triangle PAB$ is

$$G\left(\frac{x_1+6+6}{3}, \frac{y_1+2\sqrt{3}-2\sqrt{3}}{3}\right)$$ $$G\left(\frac{x_1+12}{3}, \frac{y_1}{3}\right).$$

Let the centroid be $G(x,y)$. Then

$$x_1 = 3x - 12, \quad y_1 = 3y.$$

Since $P$ lies on $2x_1 - 3y_1 + 4 = 0$, substitute these values:

$$2(3x - 12) - 3(3y) + 4 = 0$$ $$6x - 24 - 9y + 4 = 0$$ $$6x - 9y = 20.$$

Therefore, the centroid lies on the line $6x - 9y = 20$. The correct option is D.

Given: $P$ moves on the line $2x - 3y + 4 = 0$ and $A,B$ are the intersection points of the given circle and hyperbola.

Find: The locus of the centroid of $\triangle PAB$.

The key observation is that the two useful intersection points are symmetric about the $x$-axis:

$$A(6,2\sqrt{3}), \quad B(6,-2\sqrt{3}).$$

Let

$$P(x_1,y_1)$$

with

$$2x_1 - 3y_1 + 4 = 0.$$

If the centroid is $G(x,y)$, then by the centroid formula,

$$x = \frac{x_1 + 6 + 6}{3} = \frac{x_1 + 12}{3}$$

and

$$y = \frac{y_1 + 2\sqrt{3} - 2\sqrt{3}}{3} = \frac{y_1}{3}.$$

Thus,

$$x_1 = 3x - 12, \quad y_1 = 3y.$$

Substitute into the line of $P$:

$$2(3x-12) - 3(3y) + 4 = 0$$ $$6x - 24 - 9y + 4 = 0$$ $$6x - 9y = 20.$$

Hence the locus of the centroid is $6x - 9y = 20$, so the correct option is D.