Question Statement

Find the equation of the parabola in each of the following cases:

(i) Vertex at origin and focus $\left(0,-\frac{1}{32}\right)$ (ii) Vertex $(-8,-9)$ and focus $\left(-\frac{31}{4},-9\right)$ (iii) Vertex $(-6,-9)$ and directrix $x=-\frac{47}{8}$ (iv) Vertex $(5,-1)$ and $y$-intercept $-\frac{27}{2}$ (v) Focus $\left(-\frac{3}{4},-1\right)$ and directrix $y=\frac{2}{5}$ (vi) Vertex $(7,6)$, passes through $(-11,9)$, and opens left or right (vii) Opens up or down and passes through the points $(11,15),(7,7)$, and $(4,22)$ (viii) Vertex $(10,0)$, axis of symmetry $y=0$, length of latus rectum $=1$, and $a<0$ (ix) Vertex $(4,2)$, axis of symmetry $x=4$, length of latus rectum $=\frac{1}{3}$, and $a>0$ (x) Vertex at origin, opens left, and the distance between focus and vertex is $\frac{1}{8}$ units

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Background and Explanation

A parabola is the set of all points in a plane that are equidistant from a fixed point (the focus) and a fixed line (the directrix). The standard equations for a parabola with vertex $(h,k)$ are $(y-k{)}^2=\pm 4a(x-h)$ for horizontal parabolas and $(x-h{)}^2=\pm 4a(y-k)$ for vertical parabolas, where $a$ is the distance from the vertex to the focus or directrix.

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Solution

Part (i)

Given that the vertex is $V(0,0)$ and the focus is $F\left(0,-\frac{1}{32}\right)$. Since the focus is on the $y$-axis below the origin, the parabola opens downwards.

The equation of the parabola is: $x^2=-4ay$

The focus is $(0,-a)=\left(0,-\frac{1}{32}\right)$.

Comparing the coordinates, we get $a=\frac{1}{32}$. Substituting this into equation (i): $x^2=-4\left(\frac{1}{32}\right)y$ $x^2=-\frac{1}{8}y$ This is the required equation.

Part (ii)

Vertex is $(h,k)=(-8,-9)$ and focus is $F\left(-\frac{31}{4},-9\right)$. Since the focus lies to the right of the vertex (as $-\frac{31}{4}>-8$), this parabola opens to the right.

Let the equation be $(y-k{)}^2=4a(x-h)$. Substituting the vertex: $(y-(-9){)}^2=4a(x-(-8))$ $(y+9{)}^2=4a(x+8)$

The distance $a$ between the focus and the vertex is: $a=|VF|=\sqrt{{\left(-8-\left(-\frac{31}{4}\right)\right)}^2+(-9-(-9){)}^2}$ $a=\sqrt{{\left(-\frac{1}{4}\right)}^2+0^2}=\sqrt{\frac{1}{16}}=\frac{1}{4}$

Substituting $a=\frac{1}{4}$ into equation (i): $(y+9{)}^2=4\left(\frac{1}{4}\right)(x+8)$ $(y+9{)}^2=x+8$

Part (iii)

Vertex is $(h,k)=(-6,-9)$ and directrix is $x=-\frac{47}{8}$. Since the vertex is to the left of the directrix ($x=-6$ is left of $x=-5.875$), the parabola opens to the left.

Let the equation be $(y-k{)}^2=-4a(x-h)$. Substituting the vertex: $(y-(-9){)}^2=-4a(x-(-6))$ $(y+9{)}^2=-4a(x+6)$

The distance $a$ between the vertex and directrix is: $a=\frac{|8(-6)+47|}{\sqrt{8^2+0^2}}=\frac{|-48+47|}{8}=\frac{1}{8}$

Substituting $a=\frac{1}{8}$ into equation (i): $(y+9{)}^2=-4\left(\frac{1}{8}\right)(x+6)$ $(y+9{)}^2=-\frac{1}{2}(x+6)$

Part (iv)

Vertex is $(h,k)=(5,-1)$. There are four possible standard forms:

1. $(y+1{)}^2=4a(x-5)$
2. $(y+1{)}^2=-4a(x-5)$
3. $(x-5{)}^2=4a(y+1)$
4. $(x-5{)}^2=-4a(y+1)$

Given $y$-intercept is $-\frac{27}{2}$, the point $\left(0,-\frac{27}{2}\right)$ must satisfy the equation.

Case 1: $(y+1{)}^2=4a(x-5)$ ${\left(-\frac{27}{2}+1\right)}^2=4a(0-5)\Rightarrow \frac{625}{4}=-20a\Rightarrow a=-\frac{125}{16}$ Since $a$ must be positive, this case is impossible.

Case 2: $(y+1{)}^2=-4a(x-5)$ ${\left(-\frac{27}{2}+1\right)}^2=-4a(0-5)\Rightarrow \frac{625}{4}=20a\Rightarrow a=\frac{125}{16}$ Substituting $a$: $(y+1{)}^2=-4\left(\frac{125}{16}\right)(x-5)\Rightarrow (y+1{)}^2=-\frac{125}{4}(x-5)$.

Case 3: $(x-5{)}^2=4a(y+1)$ $(0-5{)}^2=4a\left(-\frac{27}{2}+1\right)\Rightarrow 25=4a\left(-\frac{25}{2}\right)\Rightarrow a=-\frac{1}{2}$ Impossible as $a<0$.

Case 4: $(x-5{)}^2=-4a(y+1)$ $(0-5{)}^2=-4a\left(-\frac{27}{2}+1\right)\Rightarrow 25=-4a\left(-\frac{25}{2}\right)\Rightarrow a=\frac{1}{2}$ Substituting $a$: $(x-5{)}^2=-4\left(\frac{1}{2}\right)(y+1)\Rightarrow (x-5{)}^2=-2(y+1)$.

Both Case 2 and Case 4 provide valid equations.

Part (v)

Focus $F\left(-\frac{3}{4},-1\right)$ and directrix $y=\frac{2}{5}$ (or $5y-2=0$).

Let $P(x,y)$ be any point on the parabola. By definition, $|PF|=|PM|$: $\sqrt{{\left(x+\frac{3}{4}\right)}^2+(y+1{)}^2}=\frac{|5y-2|}{\sqrt{0^2+5^2}}$ Squaring both sides: ${\left(x+\frac{3}{4}\right)}^2+(y+1{)}^2=\frac{(5y-2{)}^2}{25}$ $x^2+\frac{3}{2}x+\frac{9}{16}+y^2+2y+1=\frac{25y^2-20y+4}{25}$ $x^2+y^2+\frac{3}{2}x+2y+\frac{25}{16}=y^2-\frac{4}{5}y+\frac{4}{25}$ $x^2+\frac{3}{2}x+\frac{14}{5}y+\frac{561}{400}=0$

Part (vi)

Vertex $(h,k)=(7,6)$ and passes through $(-11,9)$. Since the point is to the left of the vertex, the parabola opens to the left.

Equation: $(y-k{)}^2=-4a(x-h)$ $(y-6{)}^2=-4a(x-7)$ Substitute $(-11,9)$: $(9-6{)}^2=-4a(-11-7)\Rightarrow 9=72a\Rightarrow a=\frac{1}{8}$ Substitute $a=\frac{1}{8}$ into (i): $(y-6{)}^2=-4\left(\frac{1}{8}\right)(x-7)\Rightarrow (y-6{)}^2=-\frac{1}{2}(x-7)$

Part (vii)

Let the equation be $y=a{x}^2+bx+c$. Substitute the points $(11,15),(7,7)$, and $(4,22)$:

1. $121a+11b+c=15$
2. $49a+7b+c=7$
3. $16a+4b+c=22$

Subtract (3) from (2): $72a+4b=8\Rightarrow 18a+b=2$ Subtract (4) from (3): $33a+3b=-15\Rightarrow 11a+b=-5$ Subtract (6) from (5): $7a=7\Rightarrow a=1$. Substitute $a=1$ into (6): $11(1)+b=-5\Rightarrow b=-16$. Substitute $a=1,b=-16$ into (4): $16(1)+4(-16)+c=22\Rightarrow c=70$.

Equation: $y=x^2-16x+70$ $y=(x-8{)}^2+6\Rightarrow (x-8{)}^2=y-6$

Part (viii)

Vertex $(h,k)=(10,0)$, axis $y=0$ ($x$-axis), latus rectum length $|4a|=1$. Since $a<0$, it opens left. Equation: $(y-k{)}^2=-4a(x-h)$ $(y-0{)}^2=-1(x-10)\Rightarrow y^2=-(x-10)$

Part (ix)

Vertex $(h,k)=(4,2)$, axis $x=4$ (parallel to $y$-axis), latus rectum $|4a|=\frac{1}{3}$. Since $a>0$, it opens up. Equation: $(x-h{)}^2=4a(y-k)$ $(x-4{)}^2=\frac{1}{3}(y-2)$

Part (x)

Vertex $(0,0)$, opens left, distance $a=\frac{1}{8}$. Equation: $(y-k{)}^2=-4a(x-h)$ $y^2=-4\left(\frac{1}{8}\right)x\Rightarrow y^2=-\frac{1}{2}x$

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Key Formulas or Methods Used

• Horizontal Parabola: $(y-k{)}^2=\pm 4a(x-h)$
• Vertical Parabola: $(x-h{)}^2=\pm 4a(y-k)$
• Latus Rectum Length: $L=|4a|$
• Distance Formula: $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
• Definition of Parabola: Distance from point to focus = Distance from point to directrix.

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Summary of Steps

1. Identify the vertex $(h,k)$ and the direction of the opening.
2. Determine the value of $a$ using the focus, directrix, or latus rectum length.
3. Select the appropriate standard form based on the axis of symmetry and opening direction.
4. Substitute the known values into the equation and simplify.
5. For parabolas passing through multiple points, use the general quadratic form and solve the resulting system of linear equations.