Question

Solve question using parametric coordinates method

Find the length of line segment joining the vertex of the parabola y² = 4ax and a point on the parabola where the line segment makes an angle '∅' theta to the x axis.

Can anyone also explain about the concept of parametric coordinates? my teacher solved using parametric coordinate and i didn't get that. ​

Answer 1

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

Given that,

A line segment joining the vertex of the parabola y² = 4ax and a point on the parabola, where the line segment makes an angle 'θ' to the x axis.

We know, Parametric form of y² = 4ax is x = at² , y = 2at.

Let the required point be P which lies on the parabola y² = 4ax and Let assume that coordinates of point in parametric form is P (at² , 2at)

Vertex of the parabola y² = 4ax is O(0, 0).

Now, Length of line segment OP is evaluated using distance formula as

$\rm \: OP = \sqrt{ {( {at}^{2} - 0)}^{2} + {(2at - 0)}^{2} }$

$\rm \: = \sqrt{ {( {at}^{2})}^{2} + {(2at)}^{2} }$

$\rm \: = \sqrt{ {a}^{2} {t}^{4} + 4 {a}^{2} {t}^{2} }$

$\rm \: = \sqrt{ {a}^{2} {t}^{2}( {t}^{2} + 4)}$

$\rm \: = \: at \sqrt{ {t}^{2} + 4 }$

$\rm\implies \:OP = \: at \sqrt{ {t}^{2} + 4 } - - - (1)$

Now, as line segment OP makes an angle θ, with positive direction of x axis.

So,

$\rm \: Slope \: of \: OP \: = \: tan \theta \:$

$\rm \: \dfrac{2at - 0}{ {at}^{2} - 0 } \: = \: tan \theta \:$

$\rm \: \dfrac{2at}{ {at}^{2} } \: = \: tan \theta \:$

$\rm \: \dfrac{2}{t} \: = \: tan \theta \:$

$\rm\implies \:t \: = \: 2 \: cot\theta$

So, on substituting the value of t, in equation (1), we get

$\rm \: OP = \: a(2cot\theta) \sqrt{ {(2cot\theta)}^{2} + 4 }$

$\rm \: OP = \: 2a \: cot\theta \sqrt{ {4cot^{2} \theta} + 4 }$

$\rm \: OP = \: 2a \: cot\theta \sqrt{4( {cot^{2} \theta} +1)}$

$\rm \: OP = \: 4 \: a \: cot\theta \sqrt{{cot^{2} \theta} +1}$

$\rm \: OP = \: 4 \: a \: cot\theta \sqrt{{cosec^{2} \theta}}$

$\rm\implies \:\rm \: OP = \: 4 \: a \: cot\theta \: cosec\theta$

$\rule{190pt}{2pt}$

Parametric form of Parabola

$\\ \begin{gathered}\boxed{\begin{array}{c|c} \bf Equation \: of \: parabola \: & \bf Parametric \: form \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf {y}^{2} = 4ax & \sf x = {at}^{2}, \: y = 2at \\ \\ \sf {y}^{2} = - 4ax & \sf x = - {at}^{2}, \: y = 2at \\ \\ \sf {x}^{2} = 4ay & \sf y = {at}^{2}, \: x = 2at\\ \\ \sf {x}^{2} = - 4ay & \sf y = - {at}^{2}, \: x = 2at \end{array}} \\ \end{gathered}$

Answer 2

EXPLANATION.

The length of line segment joining the vertex.

Parabola : y² = 4ax.

Making a point on the parabola where the line segment makes an angle θ to the x - axis.

As we know that,

Slope of a line joining two points (x₁, y₁) and (x₂, y₂) is (y₂ - y₁)/(x₂ - x₁).

Using this formula in the equation, we get.

Slope of line OA = (y - 0)/(x - 0).

Slope of line OA = y/x.

We know that,

Slope = tanθ.

⇒ tanθ = y/x.

⇒ y = xtanθ. - - - - - (1).

Put the value of y = xtanθ in equation of parabola, we get.

⇒ (xtanθ)² = 4ax.

⇒ x² tan²θ = 4ax.

⇒ xtan²θ = 4a.

⇒ x = 4a/tan²θ.

⇒ x = 4acot²θ.

Put the value of x = 4acot²θ in equation (1), we get.

⇒ y = (4acot²θ)(tanθ).

⇒ y = (4a/tan²θ) x tanθ.

⇒ y = 4a/tanθ.

⇒ y = 4acotθ.

Value of x = 4acot²θ  and  y = 4acotθ.

As we know that,

Distance formula.

⇒ D = √(y₂ - y₁)² + (x₂ - x₁)².

Using this formula in the equation, we get.

⇒ OA = √(y - 0)² + (x - 0)².

⇒ OA = √(y)² + (x)².

Put the values in the equation, we get.

⇒ OA = √(4acotθ)² + (4acot²θ)².

⇒ OA = √16a²cot²θ + 16a²cot⁴θ.

⇒ OA = √16a²(cot²θ + cot⁴θ).

⇒ OA = √16a²cot²θ(1 + cot²θ).

As we know that,

Formula of :

⇒ 1 + cot²θ = cosec²θ.

⇒ cot²θ = cosec²θ - 1.

Put the values in the equation, we get.

⇒ OA = √16a²cot²θ[1 + cosec²θ - 1].

⇒ OA = √16a²cot²θ(cosec²θ).

⇒ OA = 4acotθcosecθ.

Line segment joining the vertex of the parabola y² = 4ax : 4acotθcosecθ.

                                                                                                                   

MORE INFORMATION.

(1) Parametric form in straight lines.

x - x₁/cosθ = y - y₁/sinθ = r.

(2) Parametric equations of a circle.

(a) The parametric equation of a circle : x² + y² = r² are x = rcosθ and y = rsinθ.

(b) The parametric equation of the circle : (x - h)² + (y - k)² = r² are x = h + rcosθ and  y = k + rsinθ.

(c) Parametric equation of the circle : x² + y² + 2gx + 2fy + c = 0 are x = - g + √(g² + f² - c) cosθ  and  y = - f + √(g² + f² - c) sinθ.

(3) Parametric equation of parabola.

y² = 4ax are x = at² , y = 2at  and for  parabola  x² = 4ay is x = 2at  and  y = at².