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Animated Solution for Mathematics - Conic Sections: Let $P$ be the point $(1, 0)$ and $Q$ a point on the locus $y^{2} = 8 x$ . The locus of mid point of $P Q$ is

Q: Why do we introduce $h$ and $k$ instead of just using $x$ and $y$ for $Q$?

Great question! We use $h$ and $k$ to distinguish the coordinates of the moving point $Q$ from the coordinates of the locus point $M(x, y)$. If we used $x$ and $y$ for both, we would get confused. $h$ and $k$ are just placeholders that we eventually eliminate.

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Visualized Solution

Visualizing the Setup

We are given a fixed point $P (1, 0)$ on the x-axis.
A moving point $Q (h, k)$ lies on the parabola $y^{2} = 8 x$ .
Our goal is to find the path traced by the midpoint $M (x, y)$ of the line segment $P Q$ .

Defining Point $Q (h, k)$

Let the coordinates of the moving point $Q$ be $(h, k)$ .
Since $Q (h, k)$ lies on the parabola $y^{2} = 8 x$ , its coordinates must satisfy the parabola's equation.
This gives us our first relation: $k^{2} = 8 h$ .

Midpoint Formula for $M (x, y)$

Let the coordinates of the midpoint of $P Q$ be $M (x, y)$ .
Using the midpoint formula between $P (1, 0)$ and $Q (h, k)$ :
$x = \frac{h + 1}{2}$
$y = \frac{k + 0}{2}$

Expressing $h$ in terms of $x$

To find the locus, we need to eliminate the parameters $h$ and $k$ .
From the x-coordinate relation: $x = \frac{h + 1}{2}$
Multiplying by $2$ gives: $2 x = h + 1$
Rearranging for $h$ : $h = 2 x - 1$

Expressing $k$ in terms of $y$

Now let's do the same for the y-coordinate relation.
From $y = \frac{k}{2}$ :
Multiplying by $2$ gives: $k = 2 y$

Substituting $h$ and $k$

We have $h = 2 x - 1$ and $k = 2 y$ .
Substitute these expressions back into the original condition: $k^{2} = 8 h$ .
This yields: $(2 y)^{2} = 8 (2 x - 1)$

Expanding the Equation

Let's expand both sides of the equation.
On the left-hand side, $(2 y)^{2}$ becomes $4 y^{2}$ .
On the right-hand side, distributing $8$ gives: $16 x - 8$ .
So, we have: $4 y^{2} = 16 x - 8$

Simplifying the Equation

Notice that all coefficients ( $4$ , $16$ , and $8$ ) are divisible by $4$ .
Dividing the entire equation by $4$ simplifies it to:
$y^{2} = 4 x - 2$

Final Locus Equation

Rearranging all terms to one side to match the given options:
$y^{2} - 4 x + 2 = 0$
This matches option (a) perfectly!

Key Takeaways

Geometric Insight: The locus of the midpoint of a line segment from a fixed point to a parabola is another parabola.
General Strategy: Always express the coordinates of the moving point in terms of the midpoint, then substitute into the curve's equation.

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The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

We are given a fixed point

P (1, 0)

and a parabola defined by the equation

y^{2} = 8 x

. A point

Q (h, k)

moves along this parabola.

Since

Q

lies on the parabola, its coordinates must satisfy the constraint:

k^{2} = 8 h

Defining the Midpoint

Let

M (x, y)

be the midpoint of the line segment

P Q

. Using the midpoint formula, we relate the coordinates of

M

to the coordinates of

P

and

Q

x = \frac{h + 1}{2}

y = \frac{k + 0}{2}

From these equations, we can express the coordinates of

Q

in terms of the coordinates of

M

h = 2 x - 1

k = 2 y

The Master Equation

To find the locus of

M

, we substitute the expressions for

h

and

k

into the constraint equation

k^{2} = 8 h

(2 y)^{2} = 8 (2 x - 1)

Expanding the left side, we obtain:

4 y^{2} = 16 x - 8

Final Calculation

Dividing the entire equation by

4

, we arrive at the equation of the locus:

y^{2} = 4 x - 2

This result confirms that the locus of the midpoint

M

is another parabola. The transformation of the original curve through the midpoint operation preserves the parabolic nature of the path, resulting in the final equation $y^{2} = 4 x - 2$ .

Common Traps & FAQs

Why do we introduce

h

and

k

instead of just using

x

and

y

for

Q

Great question! We use

h

and

k

to distinguish the coordinates of the moving point

Q

from the coordinates of the locus point

M (x, y)

. If we used

x

and

y

for both, we would get confused.

h

and

k

are just placeholders that we eventually eliminate.

Could I have used parametric coordinates

(2 t^{2}, 4 t)

for

Q

Absolutely! That is a brilliant way to solve it. If you set

h = 2 t^{2}

and

k = 4 t

, then

x = \frac{2 t ^{2} + 1}{2}

and

y = \frac{4 t}{2} = 2 t

. Then

t = y /2

, and substituting this into the

x

equation gives

x = \frac{2 ( y /2 ) ^{2} + 1}{2} = \frac{y ^{2} /2 + 1}{2} = \frac{y ^{2} + 2}{4}

. This leads to

4 x = y^{2} + 2

, which is the same result. It's a faster path for those comfortable with parameters!

Why is the resulting locus still a parabola?

This is a fundamental property of conic sections! A linear transformation (like finding a midpoint) applied to a parabola results in another parabola. You are essentially scaling and shifting the original curve. It's beautiful, isn't it?

Animated Solution for Mathematics - Conic Sections: Let $P$ be the point $(1, 0)$ and $Q$ a point on the locus $y^{2} = 8 x$ . The locus of mid point of $P Q$ is

Select Answer:

Visualized Solution

Visualizing the Setup

Defining Point $Q (h, k)$

Midpoint Formula for $M (x, y)$

Expressing $h$ in terms of $x$

Expressing $k$ in terms of $y$

Substituting $h$ and $k$

Expanding the Equation

Simplifying the Equation

Final Locus Equation

Key Takeaways

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

Defining the Midpoint

The Master Equation

Final Calculation

Similar Questions

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Let $(x, y)$ be any point on the parabola $y^{2} = 4 x$ . Let $P$ be the point that divides the line segment from $(0, 0)$ to $(x, y)$ in the ratio $1 : 3$ . Then the locus of $P$ is

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Let $S$ be the focus of the parabola $y^{2} = 8 x$ and let $P Q$ be the common chord of the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ and the given parabola. The area of the triangle $P QS$ is

Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P be the point (1,0) and Q a point on the locus y2=8x. The locus of mid point of PQ is

Select Answer:

Visualized Solution

Visualizing the Setup

Defining Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q(h,k)

Midpoint Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M(x,y)

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Expanding the Equation

Simplifying the Equation

Final Locus Equation

Key Takeaways

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

Defining the Midpoint

The Master Equation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of a point which divides the line segment joining the point (0,−1) and a point on the parabola, x2=4y, internally in the ratio 1: 2, is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let (x,y) be any point on the parabola y2=4x. Let P be the point that divides the line segment from (0,0) to (x,y) in the ratio 1:3. Then the locus of P is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let O(0,0) and A(0,1) be two fixed points. Then the locus of a point P such that the perimeter of ΔAOP is 4, is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Given the points A(0,4) and B(0,−4), the equation of the locus of the point P(x,y) such that ∣AP−BP∣=6 is .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of the mid-point of the line segment joining the focus to a moving point on the parabola y2=4ax is another parabola with directrix

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S be the focus of the parabola y2=8x and let PQ be the common chord of the circle x2+y2−2x−4y=0 and the given parabola. The area of the triangle PQS is

Animated Solution for Mathematics - Conic Sections: Let $P$ be the point $(1, 0)$ and $Q$ a point on the locus $y^{2} = 8 x$ . The locus of mid point of $P Q$ is

Defining Point $Q (h, k)$

Midpoint Formula for $M (x, y)$

Expressing $h$ in terms of $x$

Expressing $k$ in terms of $y$

Substituting $h$ and $k$

The locus of a point which divides the line segment joining the point $(0, - 1)$ and a point on the parabola, $x^{2} = 4 y$ , internally in the ratio 1: 2, is :

Let $(x, y)$ be any point on the parabola $y^{2} = 4 x$ . Let $P$ be the point that divides the line segment from $(0, 0)$ to $(x, y)$ in the ratio $1 : 3$ . Then the locus of $P$ is

Let $O (0, 0)$ and $A (0, 1)$ be two fixed points. Then the locus of a point $P$ such that the perimeter of $Δ A O P$ is 4, is :

Given the points $A (0, 4)$ and $B (0, - 4)$ , the equation of the locus of the point $P (x, y)$ such that $∣ A P - B P ∣ = 6$ is .........

The locus of the mid-point of the line segment joining the focus to a moving point on the parabola $y^{2} = 4 a x$ is another parabola with directrix

Let $S$ be the focus of the parabola $y^{2} = 8 x$ and let $P Q$ be the common chord of the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ and the given parabola. The area of the triangle $P QS$ is