Question

find the locus of the points that are equidistant to the point F(0,2) and the line y=-2?​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

• A point is such that it is equidistant from the point F(0,2) and the line y = - 2.

Let assume that

• Required point of locus be S (x, y)

• SM be the perpendicular drop on the line y = - 2

So, According to statement

$\red{\rm :\longmapsto\:SF = SM}$

$\rm :\longmapsto\: \sqrt{ {(x - 0)}^{2} + {(y - 2)}^{2}} = \dfrac{y + 2}{ \sqrt{ {1}^{2} } }$

$\rm :\longmapsto\: \sqrt{ {x}^{2} + {(y - 2)}^{2}} = y + 2$

On squaring both sides, we get

$\rm :\longmapsto\: {x}^{2} + {(y - 2)}^{2} = {(y + 2)}^{2}$

$\rm :\longmapsto\: {x}^{2} + \cancel{y}^{2} + \cancel4 - 4y = \cancel{y}^{2} + \cancel4 + 4y$

$\rm :\longmapsto\: {x}^{2} - 4y = 4y$

$\rm :\longmapsto\: {x}^{2}= 4y + 4y$

$\rm :\longmapsto\: {x}^{2}= 8y$

This is the required locus which is the shape of upward parabola having vertex at ( 0, 0 ).

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Learn More

1. Section formula

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the point which divides PQ internally in the ratio m₁ : m₂. Then, the coordinates of R will be:

$\sf\implies R = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)$

2. Mid-point formula

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the mid-point of PQ. Then, the coordinates of R will be:

$\sf\implies R = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)$

3. Centroid of a triangle

Centroid of a triangle is the point where all the medians of the triangle meet.

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle. Let R(x, y) be the centroid of the triangle. Then, the coordinates of R will be:

$\sf\implies R = \bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3}, \: \dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg)$

Answer 2

Given:

Point F(0, 2)

Equation of a line, y=-2

To find:

the locus of the points that are equidistant from point F and the line

Solution:

According to the statements given, we can write that

$\sqrt{(h-0)^2+(k-2)^2}=\frac{|k+2|}{\sqrt{1^2} }$

⇒$\sqrt{h^2+(k-2)^2}=k+2$

On squaring both sides we get,

$h^2+(k-2)^2=(k+2)^2$

⇒$h^2=(k+2)^2-(k-2)^2$

Splitting R.H.S in the form of $a^2-b^2=(a+b)(a-b)$

⇒$h^2=(k+2+k-2)(k+2-k+2)$

⇒$h^2=(2k)(4)$

⇒$h^2=8k$

Hence the required locus of the points is  $x^2=8y$  which is an upward parabola with vertex at (0, 0).