If the point (h, k) lies on $x^{2}+y^{2}=1$, then the locus of the point $(3h+2,k)$ will be:

Question

Accepted Answer

To determine the locus of the point $(3h+2, k)$ given that the point $(h, k)$ lies on the circle $x^2 + y^2 = 1$, we need to analyze the transformation of the coordinates.

1. **Given Condition**: The point $(h, k)$ lies on the circle $x^2 + y^2 = 1$. This implies that:
   $$
   h^2 + k^2 = 1
   $$

2. **Transformation**: We are interested in the locus of the point $(3h+2, k)$. Let's denote this point as $(X, Y)$, where:
   $$
   X = 3h + 2 \quad \text{and} \quad Y = k
   $$

3. **Substitute and Simplify**: From the given condition $h^2 + k^2 = 1$, we can express $h$ in terms of $k$:
   $$
   h^2 = 1 - k^2
   $$
   Therefore, $h = \pm \sqrt{1 - k^2}$.

4. **Express $X$ in terms of $k$**:
   $$
   X = 3h + 2 = 3(\pm \sqrt{1 - k^2}) + 2
   $$

5. **Locus Equation**: To find the locus, we need to eliminate $h$ and express everything in terms of $X$ and $Y$:
   $$
   X - 2 = 3h \quad \Rightarrow \quad h = \frac{X - 2}{3}
   $$

6. **Substitute back into the circle equation**:
   $$
   \left(\frac{X - 2}{3}\right)^2 + Y^2 = 1
   $$
   Simplifying this equation:
   $$
   \frac{(X - 2)^2}{9} + Y^2 = 1
   $$
   Multiply through by 9 to clear the fraction:
   $$
   (X - 2)^2 + 9Y^2 = 9
   $$

7. **Recognize the Conic Section**: The equation $(X - 2)^2 + 9Y^2 = 9$ is in the standard form of an ellipse:
   $$
   \frac{(X - 2)^2}{9} + \frac{Y^2}{1} = 1
   $$

Thus, the locus of the point $(3h+2, k)$ is an ellipse, which confirms that Option 4 is correct.

**Explanation for Incorrect Options**:

- **Option 1 (Hyperbola)**: A hyperbola has the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ or $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$. The derived equation $(X - 2)^2 + 9Y^2 = 9$ does not match this form, as there is no subtraction between the terms.

- **Option 2 (Circle)**: A circle has the form $(x - h)^2 + (y - k)^2 = r^2$. The derived equation is not a circle because the coefficients of $X^2$ and $Y^2$ are not equal.

- **Option 3 (Parabola)**: A parabola typically has the form $y^2 = 4ax$ or $x^2 = 4ay$. The derived equation involves both $X^2$ and $Y^2$ terms, which is not characteristic of a parabola.

Therefore, the correct answer is indeed Option 4, as the locus is an ellipse.