Question Statement

Find the equation of the parabola with the focus at $(p\sin \theta ,p\cos \theta )$ and the directrix given by the equation: $x\cos \theta +y\sin \theta =p$

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Background and Explanation

A parabola is defined as the locus of all points $P(x,y)$ such that the distance from $P$ to a fixed point (the focus) is equal to the perpendicular distance from $P$ to a fixed line (the directrix). To solve this, we use the distance formula and the formula for the perpendicular distance from a point to a line.

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Solution

Let $P(x,y)$ be any point on the parabola.

By the definition of a parabola: $\text{Distance of }P\text{ from focus}=\text{Distance of }P\text{ from directrix}$

Given the focus $(p\sin \theta ,p\cos \theta )$ and the directrix $x\cos \theta +y\sin \theta -p=0$, we set up the equation: $\sqrt{(x-p\sin \theta {)}^2+(y-p\cos \theta {)}^2}=\frac{|x\cos \theta +y\sin \theta -p|}{\sqrt{{\cos }^2\theta +{\sin }^2\theta }}$

Since ${\cos }^2\theta +{\sin }^2\theta =1$, the denominator on the right side becomes $1$. Squaring both sides to remove the square root: $(x-p\sin \theta {)}^2+(y-p\cos \theta {)}^2=(x\cos \theta +y\sin \theta -p{)}^2$

Expanding both sides:

Left-Hand Side (LHS): $x^2-2px\sin \theta +p^2{\sin }^2\theta +y^2-2py\cos \theta +p^2{\cos }^2\theta$ Using the identity ${\sin }^2\theta +{\cos }^2\theta =1$, the constant terms simplify: $x^2+y^2-2px\sin \theta -2py\cos \theta +p^2$

Right-Hand Side (RHS): Using the expansion $(a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ca$: $x^2{\cos }^2\theta +y^2{\sin }^2\theta +p^2+2xy\sin \theta \cos \theta -2py\sin \theta -2px\cos \theta$

Equating and Simplifying:

Equating LHS and RHS: $x^2+y^2-2px\sin \theta -2py\cos \theta +p^2=x^2{\cos }^2\theta +y^2{\sin }^2\theta +p^2+2xy\sin \theta \cos \theta -2py\sin \theta -2px\cos \theta$

Subtract $p^2$ from both sides and group the $x^2$ and $y^2$ terms: $(x^2-x^2{\cos }^2\theta )+(y^2-y^2{\sin }^2\theta )-2xy\sin \theta \cos \theta =2px\sin \theta +2py\cos \theta -2py\sin \theta -2px\cos \theta$

Factor out the common terms: $x^2(1-{\cos }^2\theta )+y^2(1-{\sin }^2\theta )-2xy\sin \theta \cos \theta =2p\sin \theta (x-y)-2p\cos \theta (x-y)$

Apply the identity $1-{\cos }^2\theta ={\sin }^2\theta$ and $1-{\sin }^2\theta ={\cos }^2\theta$: $x^2{\sin }^2\theta +y^2{\cos }^2\theta -2xy\sin \theta \cos \theta =2p(x-y)(\sin \theta -\cos \theta )$

The left side is a perfect square $(a-b{)}^2=a^2-2ab+b^2$: $(x\sin \theta -y\cos \theta {)}^2=2p(\sin \theta -\cos \theta )(x-y)$

This is the required equation of the parabola.

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Key Formulas or Methods Used

• Parabola Definition: $PF=PM$ (Distance to focus = Distance to directrix).
• Distance Formula: $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$.
• Perpendicular Distance from $(x_1,y_1)$ to $ax+by+c=0$: $d=\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$.
• Trigonometric Identity: ${\sin }^2\theta +{\cos }^2\theta =1$.
• Algebraic Expansion: $(a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ca$.

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Summary of Steps

1. Identify the focus and the equation of the directrix.
2. Apply the definition of the parabola by setting the distance from a point $(x,y)$ to the focus equal to its distance from the directrix.
3. Square both sides of the equation to eliminate the radicals.
4. Expand both the left-hand and right-hand sides using algebraic identities.
5. Simplify the equation by using the trigonometric identity ${\sin }^2\theta +{\cos }^2\theta =1$ and canceling common terms.
6. Rearrange the remaining terms to form a perfect square for the quadratic terms and factor the linear terms to find the final equation.