Conics in General Form - Algebra 3

1. The equation $3x^2-12x+2y+16=0$ represents which conic?
   Answer: Parabola with vertex $(2,-2)$ opening downward Whenever you see a conic equation with only one squared variable, think parabola: the squared variable tells you the axis. Here $x$ is squared and $y$ is not, so it is a vertical parabola, and the sign of the $x^2$ coefficient tells you which way it opens. Solve for $y$ to put the equation in vertex form. From $$3x^2-12x+2y+16=0$$ isolate $$2y=-3x^2+12x-16$$ so $$y=-\frac32 x^2+6x-8.$$ Complete the square on $x$: $$y=-\frac32(x^2-4x)-8 = -\frac32\big((x-2)^2-4\big)-8 = -\frac32(x-2)^2+6-8 = -\frac32(x-2)^2-2.$$ In the form $y=a(x-h)^2+k$, the vertex is $(h,k)=(2,-2)$, and since $a=-\frac32<0$, the parabola opens downward. The choice with vertex $(2,2)$ comes from accidentally keeping the $+6$ from completing the square but forgetting to subtract the original $-8$, shifting the $y$-coordinate up. The choice with vertex $(-2,-2)$ opening upward reverses both the $x$-sign and the direction, as if using $(x+2)^2$ with a positive coefficient. The choice with vertex $(2,-2)$ opening upward has the correct vertex but ignores that the coefficient $a$ is negative, so the opening cannot be upward. Study tip: for a vertical parabola, always rewrite as $y=a(x-h)^2+k$. Read the vertex from $(h,k)$, then check the sign of $a$ before choosing the opening direction.
2. Find the center and radius of the circle $4x^2+4y^2-16x+8y-5=0$.
   Answer: Center $(2,-1)$, radius $\frac{5}{2}$ Divide by 4: $x^2+y^2-4x+2y-\frac{5}{4}=0$. Completing squares gives $(x-2)^2+(y+1)^2=\frac{25}{4}$, so the center is $(2,-1)$ and the radius is $\frac{5}{2}$. A radius of 5 comes from taking the square root of 25 without first dividing by 4; $(-2,1)$ reverses both center signs; $\frac{\sqrt{5}}{2}$ comes from an incorrect constant.
3. For what value of $c$ does $x^2+y^2-6x+10y+c=0$ represent a single point?
   Answer: 34 Complete the square: $(x-3)^2+(y+5)^2=34-c$. The graph is one point exactly when the right side is 0, so $34-c=0$ and $c=34$. If $c<34$ the graph is a circle; if $c>34$ there are no real points. Choice 16 uses the difference of the linear coefficients, 64 uses half of their sum squared, and 136 uses the linear coefficients without halving them.
4. For which values of $F$ does $x^2+y^2-6x+4y+F=0$ represent a real, nondegenerate circle?
   Answer: $F<13$ Complete the square: $(x-3)^2+(y+2)^2=13-F$. A real, nondegenerate circle requires the squared-radius quantity to be positive, so $13-F>0$, or $F<13$. If $F=13$, the graph collapses to a point; if $F>13$, it has no real points. The condition $F\le 13$ incorrectly includes the point-circle case.
5. After completing the square, which standard form is equivalent to $x^2-4x-8y+20=0$?
   Answer: $(x-2)^2=8(y-2)$ Complete the square in $x$: $(x-2)^2-4-8y+20=0$, so $(x-2)^2=8y-16=8(y-2)$. This is a vertical parabola with vertex $(2,2)$. A form with $(y-2)^2=8(x-2)$ would be a horizontal parabola; a negative right side would open downward; and $(x-2)^2=8(y+2)$ has the wrong vertical shift.
6. What is the graph in the real coordinate plane of $x^2-y^2-4x+6y-5=0$?
   Answer: Two intersecting lines crossing at $(2,3)$ Whenever you see a conic equation with both $x^2$ and $y^2$ terms, completing the square in each variable should be your first move. Here, $x^2-4x=(x-2)^2-4$, and $-y^2+6y=-(y-3)^2+9$. Substituting into $x^2-y^2-4x+6y-5=0$ gives $(x-2)^2-(y-3)^2=0$. This factors beautifully: $[(x-2)-(y-3)][(x-2)+(y-3)]=0$,o $x-y+1=0$ and $x+y-5=0$. So the graph is the union of two lines. Solving them together gives $x=2,\ y=3$, their intersection. Thus the graph is two intersecting lines crossing at $(2,3)$. Why not the other choices? A hyperbola centered at $(2,3)$ would appear if the completed equation were $(x-2)^2-(y-3)^2=\text{positive constant}$. The opposite signs on the squared terms do suggest a hyperbola, but here the right-hand side is $0$, and the equation degenerates into a pair of lines. The parabola with vertex $(2,3)$ would require something like $(y-3)^2=4(x-2)$, not a difference of squares. The ellipse centered at $(2,3)$ would require both squared terms to have the same positive sign, not opposite signs. So those all miss the key feature: $A^2-B^2=0$ factors into two lines. Study takeaway: when completing the square in a conic equation, always check the constant on the right-hand side. If you see $A^2-B^2=0$, factor immediately—the graph is two intersecting lines. If that constant is nonzero, then it is a hyperbola.
7. For a positive value of $a$, the equation $x^2+y^2-ax+6y+23=0$ has radius 2. What is $a$?
   Answer: $6\sqrt{2}$ Complete the square: $\left(x-\frac{a}{2}\right)^2+(y+3)^2=\frac{a^2}{4}+9-23=\frac{a^2}{4}-14$. Since the radius is 2, the right side is 4: $\frac{a^2}{4}-14=4$, so $a^2=72$ and $a=6\sqrt{2}$. The other choices come from using the wrong constant term or solving $\frac{a^2}{4}=2$ or $\frac{a^2}{4}=32$.
8. How many real points lie on the graph of $x^2+y^2+2x-4y+6=0$?
   Answer: 0 Completing the square gives $(x+1)^2-1+(y-2)^2-4+6=0$, so $(x+1)^2+(y-2)^2=-1$. Since the sum of two squares cannot be negative, no real ordered pair satisfies the equation. A zero right side would give a single point, and a positive right side would give a circle with infinitely many real points; a single nondegenerate conic graph cannot consist of exactly two points.
9. Which statement correctly classifies the graph of $4x^2+9y^2-16x+18y-11=0$?
   Answer: It is an ellipse with center $(2,-1)$ and horizontal major axis of length 6. Group terms: $4(x^2-4x)+9(y^2+2y)-11=0$. Completing squares gives $4[(x-2)^2-4]+9[(y+1)^2-1]-11=0$, so $4(x-2)^2+9(y+1)^2=36$, or $\frac{(x-2)^2}{9}+\frac{(y+1)^2}{4}=1$. The coefficients have the same sign but are unequal, so the conic is an ellipse, not a circle or hyperbola. The center is $(2,-1)$ and the larger denominator is under $x$, so the horizontal major axis has length $2(3)=6$. The other choices have incorrect signs, equal coefficients, or opposite signs.
10. Which equation represents an ellipse with center $(-1,2)$ and a vertical major axis?
    Answer: $9x^2+4y^2+18x-16y-11=0$ Complete the square for choice D: $9(x^2+2x)+4(y^2-4y)-11=0$ becomes $9(x+1)^2+4(y-2)^2=36$, or $\frac{(x+1)^2}{4}+\frac{(y-2)^2}{9}=1$. The center is $(-1,2)$, and the larger denominator is under $y$, so the major axis is vertical. Choice B has the correct center but the larger denominator under $x$, making the major axis horizontal; choices A and C have mismatched linear terms and do not have that center.
11. What is the radius of the circle represented by $2x^2+2y^2-12x+8y+20=0$?
    Answer: $\sqrt{3}$ Whenever a circle equation is given in expanded general form, your first move should be to complete the square to rewrite it as $(x-h)^2+(y-k)^2=r^2$. Here, divide every term by 2 first: $x^2+y^2-6x+4y+10=0$. Group the $x$ and $y$ terms and complete the square: $(x^2-6x+9)+(y^2+4y+4)=-10+9+4$. This gives $(x-3)^2+(y+2)^2=3$. In standard form, the right side is $r^2$, so $r^2=3$ and the radius is $r=\sqrt{3}$. The trap choices come from confusing $r^2$ with $r$. The choice $3$ is the value of $r^2$, not the radius itself. The choice $9$ may tempt you if you square $3$ or mistake the constant for the radius. The choice $\sqrt{10}$ appears if you forget to move the constant correctly and treat $10$ as $r^2$ after completing the square; here the completed-square right side is $3$, not $10$. Study tip: always get the equation into $(x-h)^2+(y-k)^2=r^2$, then take the square root of the right side. On this exam, the most common circle trap is reporting $r^2$ instead of $r$, so check that final step.
12. The equation $4x^2+ky^2-8x+12y-7=0$ represents a circle. Which statement gives the correct value of $k$ and the circle's radius?
    Answer: $k=4$, radius $\sqrt{5}$ A circle requires equal coefficients of $x^2$ and $y^2$, so $k=4$. Dividing by 4 gives $x^2+y^2-2x+3y-\frac{7}{4}=0$. Completing the square yields $(x-1)^2+\left(y+\frac{3}{2}\right)^2=5$, so the radius is $\sqrt{5}$. Choice A forgets to take the square root; choice B fails to divide by the common leading coefficient; choice C uses opposite signs, which would not be a circle.
13. After completing the square, which standard form is equivalent to $y^2-9x^2+4y-18x-14=0$?
    Answer: $\frac{(y+2)^2}{9}-(x+1)^2=1$ Group as $(y^2+4y)-9(x^2+2x)-14=0$. Completing squares gives $(y+2)^2-4-9[(x+1)^2-1]-14=0$, so $(y+2)^2-9(x+1)^2=9$. Dividing by 9 yields $\frac{(y+2)^2}{9}-(x+1)^2=1$. A form with $\frac{(x+1)^2}{9}-(y+2)^2=1$ reverses the transverse axis; $(y+2)^2-\frac{(x+1)^2}{9}=1$ is not the result of dividing by 9; and replacing the minus sign with a plus would make an ellipse.
14. The graph of $x^2+y^2-8x+2y+17=0$ is best described as which of the following?
    Answer: A single point $(4,-1)$. Complete the square: $(x-4)^2-16+(y+1)^2-1+17=0$, so $(x-4)^2+(y+1)^2=0$. A sum of squares equal to 0 is satisfied only at $x=4,\ y=-1$. Thus the degenerate conic is a single point. Choice A incorrectly treats the center coordinate 4 as the radius; choice B has sign errors; choice D would require the right side to be negative.
15. A student divides $2x^2+2y^2-8x+4y+8=0$ by 2 and claims the graph is a circle of radius 2 centered at $(2,-1)$. What is the actual graph?
    Answer: A circle centered at $(2,-1)$ with radius 1 When you see an equation like $2x^2+2y^2-8x+4y+8=0$, your first instinct should be to check whether it can be written in the standard circle form $(x-h)^2+(y-k)^2=r^2$. Because the coefficients of $x^2$ and $y^2$ are equal, this is indeed a circle—but you must complete the square to find its true center and radius. Divide the whole equation by 2 first: $x^2+y^2-4x+2y+4=0$. Now group and complete the square: $(x^2-4x)+(y^2+2y)+4=0$ $(x-2)^2-4+(y+1)^2-1+4=0$ $(x-2)^2+(y+1)^2=1$. So the actual graph is a circle centered at $(2,-1)$ with radius $\sqrt{1}=1$. The student's claim of radius 2 likely came from mistakenly treating the constant 4 (or the coefficient of the linear term) as the squared radius. A radius of 2 would require $r^2=4$ on the right, but our completed square gives 1. The center at $(-2,1)$ would arise if you incorrectly used $(x+2)^2$ and $(y-1)^2$, but the signs here are negative for $x$ and positive for $y$, so the center is indeed $(2,-1)$. Finally, "no real points" would be true only if the right side were negative (a negative radius squared), but here it is positive 1, so the circle exists. Your takeaway: never guess the radius from the constants before completing the square. Always rewrite in standard form, and double-check the signs inside the parentheses to avoid flipping the center.
16. After completing the square, which standard form represents $9y^2-4x^2+18y+16x-43=0$?
    Answer: $\frac{(y+1)^2}{4}-\frac{(x-2)^2}{9}=1$ Whenever you see a conic in expanded form, your first thought should be: complete the square for each variable to see which standard shape hides inside. Here, the equation has both $y^2$ and $x^2$ with opposite signs, so it is a hyperbola. Group the $y$-terms and $x$-terms separately. For $y$: $9y^2+18y=9(y^2+2y)=9[(y+1)^2-1]=9(y+1)^2-9$. For $x$: $-4x^2+16x=-4(x^2-4x)=-4[(x-2)^2-4]=-4(x-2)^2+16$. Substitute back: $9(y+1)^2-9-4(x-2)^2+16-43=0$. Simplify the constants: $-9+16-43=-36$, so $9(y+1)^2-4(x-2)^2-36=0$. Move the constant: $9(y+1)^2-4(x-2)^2=36$. Divide every term by $36$: $\frac{(y+1)^2}{4}-\frac{(x-2)^2}{9}=1$. That is the correct standard form. The choice with $\frac{(x-2)^2}{9}-\frac{(y+1)^2}{4}=1$ reverses which term is positive, flipping the hyperbola's orientation. The choice with $(y-1)^2$ and $(x+2)^2$ uses the wrong signs inside the binomials, likely from misreading $y+1$ as $y-1$ or $x-2$ as $x+2$. The choice with $\frac{(y+1)^2}{9}-\frac{(x-2)^2}{4}=1$ swaps the denominators: after dividing by $36$, the $y$-coefficient $9$ gives $1/4$, not $1/9$. Your strategy: always complete the square separately for each variable, factor out the leading coefficient first, then divide by the constant to get $1$. Double-check the binomial signs and denominators—they are the most common traps.
17. For what real value of $k$ does $(k-1)x^2+(k^2-1)y^2+4x-8y+5=0$ represent a parabola?
    Answer: $-1$ A parabola in this setting has exactly one nonzero quadratic term. Setting $k^2-1=0$ gives $k=\pm1$. If $k=1$, then $k-1=0$ also, so both quadratic terms vanish and the equation becomes the line $4x-8y+5=0$. Only $k=-1$ leaves the nonzero quadratic term $-2x^2$, giving a parabola. Choice D is the line trap; choices A and C leave two nonzero squared terms.
18. Which statement correctly classifies the graph of $9x^2-16y^2+36x+32y-124=0$?
    Answer: Hyperbola with center $(-2,1)$ and horizontal transverse axis. Group and complete squares: $9[(x+2)^2-4]-16[(y-1)^2-1]-124=0$, so $9(x+2)^2-16(y-1)^2=144$. Dividing by 144 gives $\frac{(x+2)^2}{16}-\frac{(y-1)^2}{9}=1$. Opposite signs mean a hyperbola, the center is $(-2,1)$, and the positive $x$-term makes the transverse axis horizontal. Choice B reverses both the center signs and the axis orientation; choice C treats opposite signs as an ellipse; choice D has the wrong axis orientation.
19. What is the graph in the real coordinate plane of $x^2+y^2-6x+4y+13=0$?
    Answer: A single point located at the coordinate $(3,-2)$ Whenever you see a quadratic in both $x$ and $y$ with equal squared coefficients, complete the square in $x$ and $y$ separately. That tells you exactly what shape the equation represents. . Group: $(x^2-6x)+(y^2+4y)+13=0$. Completing squares, $x^2-6x=(x-3)^2-9$ and $y^2+4y=(y+2)^2-4$. Substituting gives $(x-3)^2+(y+2)^2-9-4+13=0$,so $(x-3)^2+(y+2)^2=0$. Since a sum of squares equals zero only when each square is zero, $x=3$ and $y=-2$. The graph is exactly the single point $(3,-2)$. The choice saying "a circle centered at $(3,-2)$ with radius 1" treats the right side as $1$, but it is actually $0$; the radius would be $\sqrt{0}=0$. The choice saying "a circle centered at $(-3,2)$ with radius 1" makes sign errors in the center and invents a radius: it would correspond to $(x+3)^2+(y-2)^2=1$. The choice saying "no real points" is tempting because a sum of squares might seem impossible, but it equals $0$ at exactly one real point, so the graph is not empty. Strategy: after completing the square, compare the right side to zero: positive gives a circle, zero gives a single point, negative gives no real points. This pattern turns a tricky conic question into a quick check.
20. What is the graph in the real coordinate plane of $x^2+y^2-6x+4y+14=0$?
    Answer: No real points in the plane (the empty set) Whenever you see an equation with $x^2$ and $y^2$ having equal coefficients, the graph is a circle, a point, or nothing at all. Your job is to complete the square on both variables to find the standard form. Group the $x$-terms and $y$-terms: $$(x^2-6x)+(y^2+4y)+14=0.$$ Completing the squares gives $$(x-3)^2-9+(y+2)^2-4+14=0,$$ so $$(x-3)^2+(y+2)^2=-1.$$ Since a sum of two squares cannot be negative, there are no real points satisfying the equation. The graph is the empty set. The distractor "circle centered at $(3,-2)$ with radius 1" would require the right side to be $1$, but here it is $-1$. The distractor"circle centered at $(3,-2)$ with radius $\sqrt{14}$" mistakenly treats the original constant $14$ as the squared radius, but completing the square shows the right side is actually negative. The distractor"single point located at $(3,-2)$" would require $$(x-3)^2+(y+2)^2=0,$$ which would happen if the constant were $13$ instead of $14$; with $14$, the equation collapses to no real solution. Study tip: after completing the square, always check the sign of the number on the right. If it is positive, you have a circle with radius $\sqrt{\text{that number}}$; if zero, a single point; if negative, no real points. Don't mistake the original constant term fora radius—complete the square first.

Start this quiz