Let P (x, y) be the midpoint of the line joining (1, 0) to the point on the curve y^2= (x+1)(x+5) – (x+2)(x+3). The locus of P is symmetric about ?

a). y-axis
b). x=1
c). y=1
d). x-axis
e). (0, 0)

Question

a). y-axis
b). x=1
c). y=1
d). x-axis
e). (0, 0)

Caroline Aucoin · Accepted Answer

See below.

Note that #y^2=(x + 1) (x + 5) - (x + 2) (x + 3) = x-1#

so the curve is the parabola

#y^2=x-1#

The mid point locus #P(x,y) = 1/2(x+1,y)# is symmetric about the #x# axis

See attached plot.

In red #y^2= x-1# and in blue #P(x,y) = 1/2(x+1,y)#

Carson Arnold · Answer

undefined To find the locus of the midpoint $P(x, y)$ for the given curve and conditions, first, let's simplify the given curve equation:

$$
y^2 = (x+1)(x+5) - (x+2)(x+3)
$$

$$
y^2 = x^2 + 6x + 5 - (x^2 + 5x + 6)
$$

$$
y^2 = x
$$

Now, let $Q(x_1, y_1)$ be a point on the curve $y^2 = x$, hence $y_1^2 = x_1$.

The midpoint $P$ of the line joining the points $(1, 0)$ and $Q(x_1, y_1)$ is given by:

$$
x = \frac{1 + x_1}{2}, \quad y = \frac{0 + y_1}{2}
$$

Solving for $x_1$ and $y_1$ in terms of $x$ and $y$:

$$
x_1 = 2x - 1, \quad y_1 = 2y
$$

Since $y_1^2 = x_1$, we substitute $2y$ for $y_1$ and $2x - 1$ for $x_1$:

$$
(2y)^2 = 2x - 1
$$

$$
4y^2 = 2x - 1
$$

$$
2x = 4y^2 + 1
$$

$$
x = 2y^2 + \frac{1}{2}
$$

The equation $x = 2y^2 + \frac{1}{2}$ represents a parabola that is symmetric about the $y$-axis, as it can be rewritten in the form $x - \frac{1}{2} = 2y^2$, indicating the vertex at $\left(\frac{1}{2}, 0\right)$ and opening to the right. The locus of $P$ is thus symmetric about the $y$-axis.

Let P (x, y) be the midpoint of the line joining (1, 0) to the point on the curve y^2= (x+1)(x+5) – (x+2)(x+3). The locus of P is symmetric about ?

a). y-axis b). x=1 c). y=1 d). x-axis e). (0, 0)

a). y-axis
b). x=1
c). y=1
d). x-axis
e). (0, 0)