If A and B are the points of intersection of the circle x^2 + y^2 - 8x = 0 and the hyperbola \fracx^29 - \fracy^24 = 1 and a point P moves on the line 2x - 3y + 4 = 0 , then the centroid of \Delta PAB lies on the line :

Straight Lines and Pair of Straight Lines PYQ — JEE Main 2025 · 28 Jan · Shift 2 · Q37 with Solution

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Correct answer: D

Find the fixed points $A$ and $B$

We are given: $x^2+y^2-8x=0$ and $\frac{x^2}{9}-\frac{y^2}{4}=1$

First, rewrite the hyperbola equation: $\frac{x^2}{9}-\frac{y^2}{4}=1 \implies 4x^2-9y^2=36$

From the circle, $y^2=8x-x^2$

Substitute into the hyperbola: $4x^2-9(8x-x^2)=36$ $4x^2-72x+9x^2=36$ $13x^2-72x-36=0$

Solve this quadratic: $13x^2-72x-36=0$ $x=\frac{72\pm\sqrt{72^2+4\cdot 13\cdot 36}}{26}$ $=\frac{72\pm\sqrt{5184+1872}}{26}$ $=\frac{72\pm\sqrt{7056}}{26}$ $=\frac{72\pm 84}{26}$

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

Now check which values satisfy the circle:

If $x=6$ , $y^2=8(6)-36=12 \implies y=\pm 2\sqrt{3}$
If $x=-\frac{6}{13}$ , $y^2=8\left(-\frac{6}{13}\right)-\frac{36}{169}<0$ impossible.

Hence the intersection points are $A=(6,2\sqrt{3}), \quad B=(6,-2\sqrt{3})$

Find the centroid of $\triangle PAB$

Let $P=(x,y)$ where $P$ moves on the line $2x-3y+4=0$

The centroid $G$ of triangle $PAB$ is

=\left(\frac{x+12}{3},\frac{y}{3}\right)$$ Let centroid coordinates be $(h,k)$. Then $$h=\frac{x+12}{3}, \qquad k=\frac{y}{3}$$ So, $$x=3h-12, \qquad y=3k$$ 3. **Use the condition that $P$ lies on the given line** Since $P$ lies on $$2x-3y+4=0$$ substitute $x=3h-12$ and $y=3k$: $$2(3h-12)-3(3k)+4=0$$ $$6h-24-9k+4=0$$ $$6h-9k-20=0$$ Thus the centroid lies on the line $$6h-9k=20$$ Replacing $(h,k)$ by $(x,y)$, the required locus is $$\boxed{6x-9y=20}$$ 4. **Match with options** This is option **D**.

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View written solutionFree

Correct answer: D

Find the fixed points $A$ and $B$

We are given: $x^2+y^2-8x=0$ and $\frac{x^2}{9}-\frac{y^2}{4}=1$

First, rewrite the hyperbola equation: $\frac{x^2}{9}-\frac{y^2}{4}=1 \implies 4x^2-9y^2=36$

From the circle, $y^2=8x-x^2$

Substitute into the hyperbola: $4x^2-9(8x-x^2)=36$ $4x^2-72x+9x^2=36$ $13x^2-72x-36=0$

Solve this quadratic: $13x^2-72x-36=0$ $x=\frac{72\pm\sqrt{72^2+4\cdot 13\cdot 36}}{26}$ $=\frac{72\pm\sqrt{5184+1872}}{26}$ $=\frac{72\pm\sqrt{7056}}{26}$ $=\frac{72\pm 84}{26}$

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

Now check which values satisfy the circle:

If $x=6$ , $y^2=8(6)-36=12 \implies y=\pm 2\sqrt{3}$
If $x=-\frac{6}{13}$ , $y^2=8\left(-\frac{6}{13}\right)-\frac{36}{169}<0$ impossible.

Hence the intersection points are $A=(6,2\sqrt{3}), \quad B=(6,-2\sqrt{3})$

Find the centroid of $\triangle PAB$

Let $P=(x,y)$ where $P$ moves on the line $2x-3y+4=0$

The centroid $G$ of triangle $PAB$ is

=\left(\frac{x+12}{3},\frac{y}{3}\right)$$ Let centroid coordinates be $(h,k)$. Then $$h=\frac{x+12}{3}, \qquad k=\frac{y}{3}$$ So, $$x=3h-12, \qquad y=3k$$ 3. **Use the condition that $P$ lies on the given line** Since $P$ lies on $$2x-3y+4=0$$ substitute $x=3h-12$ and $y=3k$: $$2(3h-12)-3(3k)+4=0$$ $$6h-24-9k+4=0$$ $$6h-9k-20=0$$ Thus the centroid lies on the line $$6h-9k=20$$ Replacing $(h,k)$ by $(x,y)$, the required locus is $$\boxed{6x-9y=20}$$ 4. **Match with options** This is option **D**.

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