Let $A(a,0)$ and $B(-a,0)$ be two fixed points. The locus of variable point P which moves in such a way that $(PA)^{2}-(PB)^{2}=4a^{2}$ is:

Q: Let $A(a,0)$ and $B(-a,0)$ be two fixed points. The locus of variable point P which moves in such a way that $(PA)^{2}-(PB)^{2}=4a^{2}$ is:

To determine the locus of the point $ P(x, y) $ such that $(PA)^2 - (PB)^2 = 4a^2$, where $ A(a, 0) $ and $ B(-a, 0) $ are fixed points, we need to analyze the given condition. 1. **Explanation for Option 1 (Straight Line):** The expression $(PA)^2 - (PB)^2 = 4a^2$ can be expanded using the distance formula: \[ (PA)^2 = (x - a)^2 + y^2 \] \[ (PB)^2 = (x + a)^2 + y^2 \] Substituting these into the given condition: \[ [(x - a)^2 + y^2] - [(x + a)^2 + y^2] = 4a^2 \] Simplifying, we get: \[ (x^2 - 2ax + a^2 + y^2) - (x^2 + 2ax + a^2 + y^2) = 4a^2 \] \[ -4ax = 4a^2 \] Dividing both sides by $-4a$ (assuming $a \neq 0$): \[ x = -a \] This equation represents a vertical line $x = -a$, which is a straight line. Therefore, the locus of point $ P $ is indeed a straight line, confirming that Option 1 is correct. 2. **Explanation for why other options are incorrect:** - **Option 2 (Circle):** A circle is defined by the equation $(x - h)^2 + (y - k)^2 = r^2$. The derived equation $x = -a$ does not fit this form, as it lacks a $y$ term and does not represent a set of points equidistant from a center point. - **Option 3 (Parabola):** A parabola typically has the form $y^2 = 4ax$ or $x^2 = 4ay$. The equation $x = -a$ is a vertical line, not a curve opening in any direction, and thus does not match the form of a parabola. - **Option 4 (Ellipse):** An ellipse has the general form $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$. The equation $x = -a$ is a straight line and does not fit the form of an ellipse, which involves both $x$ and $y$ terms and represents a closed curve. In conclusion, the correct answer is Option 1, as the locus of point $ P $ is a straight line given by $x = -a$.

Question

Accepted Answer

To determine the locus of the point $ P(x, y) $ such that $(PA)^2 - (PB)^2 = 4a^2$, where $ A(a, 0) $ and $ B(-a, 0) $ are fixed points, we need to analyze the given condition.

1. **Explanation for Option 1 (Straight Line):**

The expression $(PA)^2 - (PB)^2 = 4a^2$ can be expanded using the distance formula:
   $$
   (PA)^2 = (x - a)^2 + y^2
   $$
   $$
   (PB)^2 = (x + a)^2 + y^2
   $$
   Substituting these into the given condition:
   $$
   [(x - a)^2 + y^2] - [(x + a)^2 + y^2] = 4a^2
   $$
   Simplifying, we get:
   $$
   (x^2 - 2ax + a^2 + y^2) - (x^2 + 2ax + a^2 + y^2) = 4a^2
   $$
   $$
   -4ax = 4a^2
   $$
   Dividing both sides by $-4a$ (assuming $a \neq 0$):
   $$
   x = -a
   $$
   This equation represents a vertical line $x = -a$, which is a straight line. Therefore, the locus of point $ P $ is indeed a straight line, confirming that Option 1 is correct.

2. **Explanation for why other options are incorrect:**

- **Option 2 (Circle):** A circle is defined by the equation $(x - h)^2 + (y - k)^2 = r^2$. The derived equation $x = -a$ does not fit this form, as it lacks a $y$ term and does not represent a set of points equidistant from a center point.

- **Option 3 (Parabola):** A parabola typically has the form $y^2 = 4ax$ or $x^2 = 4ay$. The equation $x = -a$ is a vertical line, not a curve opening in any direction, and thus does not match the form of a parabola.

- **Option 4 (Ellipse):** An ellipse has the general form $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$. The equation $x = -a$ is a straight line and does not fit the form of an ellipse, which involves both $x$ and $y$ terms and represents a closed curve.

In conclusion, the correct answer is Option 1, as the locus of point $ P $ is a straight line given by $x = -a$.