Conic Sections Practice Problems with Solutions

The standard equation of a circle centered at the origin (0, 0) is \(x^2+y^2=r^2,\) where r is the radius. Since the diameter is \(2\sqrt{11},\) the radius is half of this value:

$$r=\dfrac{2\sqrt{11}}{2}=\sqrt{11}$$

Substituting into the equation, we get:

$$x^2+y^2=(\sqrt{11})^2 → x^2+y^2=11$$

For a circle with center at \((h, k)=(3,-4)\) and radius \(r=6,\) the standard equation is:

$$(x-h)^2+(y-k)^2=r^2$$

Substituting the given values:

$$(x-3)^2+(y+4)^2=6^2$$

$$(x-3)^2+(y+4)^2=36$$

Given values:

• Center: (0, 0)
• Focus: (0, 3)
• Semi-major axis: a = 5

Since the center and focus lie on the y-axis, this is a vertical ellipse, which means its standard equation is of the form:

$$\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$$

We know one semi-axis; we can determine the other using the relationship between \(a, b,\) and \(c\). First, we find the semi-focal distance \(c,\) which is the distance between the center and the focus; in this case, \(c = 3\).

Now, we apply the relationship for an ellipse:

$$c^2=a^2-b^2→b^2=a^2-c^2$$

Thus:

$$b^2=5^2-3^2=25-9=16$$

With these values, we can write the standard form equation of the ellipse:

$$\dfrac{x^2}{16}+\dfrac{y^2}{25}=1$$

Knowing the values of \(a, b,\) and \(c,\) we can determine all other features of the ellipse.

Key Features of the Ellipse:

• Foci: (0, 3) and (0, -3)
• Center: (0, 0)
• Semi-axes: a = 5, b = 4
• Focal distance: 2c = 6
• Vertices: (0, 5) and (0, -5)
• Co-vertices: (4, 0) and (-4, 0)
• Latus rectum: \(L_R=\dfrac{2b^2}{a}=\dfrac{2\cdot 16}{5}=6.4\)
• Eccentricity: \(e=\dfrac{c}{a}=\dfrac{3}{5}=0.6\)

Given values:

• Center: (0, 0)
• Vertex: (0, 5)
• Co-vertex: (3, 0)
• Major axis along the y-axis.

The major axis lies on the y-axis, so the standard equation of the ellipse is:

$$\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$$

where \(a\) is the semi-major axis and \(b\) is the semi-minor axis.

The vertex at (0, 5) indicates that the length of the semi-major axis is \(a = 5\).

The co-vertex at (3, 0) indicates that the length of the semi-minor axis is \(b = 3\).

We use the fundamental relationship for ellipses:

$$c^2=a^2-b^2$$

Substituting the values:

$$c^2=5^2-3^2=25-9=16$$

Therefore:

$$c=\sqrt{16}=4$$

With \(a^2=25\) and \(b^2=9,\) the standard equation of the ellipse is:

$$\dfrac{x^2}{9}+\dfrac{y^2}{25}=1$$

Key Features of the Ellipse:

• Center: (0, 0)
• Foci: (0, 4) and (0, -4), since c = 4 and they lie on the y-axis.
• Semi-axes: a = 5, b = 3
• Vertices: (0, 5) and (0, -5)
• Co-vertices: (3, 0) and (-3, 0)
• Focal distance: 2c = 8
• Latus rectum: \(L_R=\dfrac{2b^2}{a}=\dfrac{2\cdot9}{5}=\dfrac{18}{5}=3.6\)
• Eccentricity: \(e=\dfrac{c}{a}=\dfrac{4}{5}=0.8\)

Given Data:

• Center of the hyperbola: C (4, -1)
• Focus: (7, -1)
• Vertex: (6, -1)

We observe that the center, focus, and vertex all share the same y-coordinate, which indicates that the hyperbola is horizontally oriented. Therefore, the standard equation of the hyperbola is:

$$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$$

where \((h, k)\) is the center of the hyperbola.

The center is C (4, -1), so \(h=4\) and \(k=-1\).

The distance between the center and the vertex is \(a\). Since the vertex is (6, -1):

$$a=6-4=2$$

The distance between the center and the focus is \(c\). Since the focus is (7, -1):

$$c=7-4=3$$

For a hyperbola, the relation between the parameters is:

$$c^2=a^2+b^2$$

Substituting the known values:

$$3^2=2^2+b^2 → 9=4+b^2 → b^2=5 → b=\sqrt{5}$$

Substituting \(h, k, a,\) and \(b\) into the equation:

$$\dfrac{(x-4)^2}{4}-\dfrac{(y+1)^2}{5}=1$$

For a horizontal hyperbola, the equations of the asymptotes are:

$$y-k=\pm \dfrac{b}{a}(x-h)$$

Substituting the values:

$$y+1=\pm \dfrac{\sqrt{5}}{2}(x-4)$$

The eccentricity \(e\) of a hyperbola is given by:

$$e=\dfrac{c}{a}$$

Substituting the known values:

$$e=\dfrac{3}{2}=1.5$$

The length of the latus rectum \(L_R\) is calculated as:

$$L_R=\dfrac{2b^2}{a}$$

Substituting the known values:

$$L_R=\dfrac{2 \cdot 5}{2}=5$$

Key Features of the Hyperbola:

• Center: (4, -1)
• Foci: (7, -1) and (1, -1)
• Vertices: (6, -1) and (2, -1)
• Semi-transverse axis: \(a=2\)
• Semi-conjugate axis: \(b=\sqrt{5}\)
• Asymptotes: \(y+1=\pm \dfrac{\sqrt{5}}{2}(x-4)\)
• Eccentricity: \(e=1.5\)
• Latus rectum: \(L_R=5\)

Equation: \(\dfrac{(x-4)^2}{4}-\dfrac{(y+1)^2}{5}=1\)

Given Data:

• Foci: (3, 7) and (7, 7)
• Vertex: (6, 7)

We observe that both foci and the vertex share the same y-coordinate (\(y = 7\)), which indicates that the hyperbola is horizontally oriented. Therefore, its standard equation is:

$$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$$

where \((h, k)\) is the center of the hyperbola.

The center is the midpoint between the foci. Given the foci (3, 7) and (7, 7), the center is calculated as:

$$h=\dfrac{3+7}{2}=5, \quad k=\dfrac{7+7}{2}=7$$

Thus, the center is C (5, 7).

The distance between the center and the vertex is \(a\). With vertex (6, 7) and center (5, 7):

$$a=6-5=1$$

The distance between the center and one of the foci is \(c\). With focus (7, 7) and center (5, 7):

$$c=7-5=2$$

For a hyperbola, \(c^2=a^2+b^2.\) Substituting the known values:

$$2^2=1^2+b^2 → 4=1+b^2 → b^2=3 → b=\sqrt{3}$$

Substituting \(h=5, k=7, a=1,\) and \(b=\sqrt{3}\) into the equation, we obtain:

$$\dfrac{(x-5)^2}{1}-\dfrac{(y-7)^2}{3}=1$$

Simplifying, the equation becomes:

$$(x-5)^2-\dfrac{(y-7)^2}{3}=1$$

For a horizontal hyperbola, the equations of the asymptotes are:

$$y-k=\pm \dfrac{b}{a}(x-h)$$

Substituting the known values:

$$y-7=\pm \dfrac{\sqrt{3}}{1}(x-5) → y-7=\pm \sqrt{3}(x-5)$$

The eccentricity of the hyperbola is:

$$e=\dfrac{c}{a}=\dfrac{2}{1}=2$$

The length of the latus rectum is:

$$L_R=\dfrac{2b^2}{a}=\dfrac{2 \cdot 3}{1}=6$$

Key Features of the Hyperbola:

• Center: (5, 7)
• Foci: (7, 7) and (3, 7)
• Vertices: (6, 7) and (4, 7)
• Semi-transverse axis: \(a=1\)
• Semi-conjugate axis: \(b=\sqrt{3}\)
• Asymptotes: \(y-7=\pm \sqrt{3}(x-5)\)
• Eccentricity: \(e=2\)
• Latus rectum: \(L_R=6\)

Equation: \((x-5)^2-\dfrac{(y-7)^2}{3}=1\)

First, identify the coordinates of the vertex \(V(3, 1)\) and focus \(F(3, -1)\). Both points share the same x-coordinate, which means the axis of symmetry is vertical. We calculate the parameter \(p\) as the difference between the y-coordinates of the focus and the vertex:

$$p=-1-1=-2$$

Since \(p\) is negative, the parabola opens downward (which is also evident because the focus lies below the vertex). Using the standard form for a vertical parabola:

$$(x-h)^2=4p (y-k)$$

where \((h, k)\) is the vertex, we substitute the known values:

$$(x-3)^2=4 (-2)(y-1)$$

Simplifying, we obtain the equation:

$$(x-3)^2=-8 (y-1)$$

Here, the vertex is at \(V (-4, 3)\) and the focus is at \(F (-1, 3)\). Both points have the same y-coordinate, so the axis of symmetry is horizontal. We determine the parameter \(p\) as the difference between the x-coordinates of the focus and vertex:

$$p=-1-(-4)=3$$

Since \(p\) is positive, the parabola opens to the right. The standard form for a horizontal parabola is:

$$(y-k)^2=4p (x-h)$$

Substituting the given coordinates:

$$(y-3)^2=4 (3)(x-(-4))$$

Simplifying, we arrive at:

$$(y-3)^2=12 (x+4)$$

We begin by analyzing the equation \(x^2+y^2-2x-2y=0.\) It contains quadratic terms in both \(x\) and \(y\) with equal coefficients for \(x^2\) and \(y^2,\) suggesting a circle. To confirm, we complete the square for both variables.

Group the \(x\) and \(y\) terms:

$$x^2-2x+y^2-2y=0$$

Complete the square for \(x\):

$$x^2-2x=(x^2-2x+1)-1=(x-1)^2-1$$

Do the same for \(y\):

$$y^2-2y=(y^2-2y+1)-1=(y-1)^2-1$$

Substitute back into the equation:

$$(x-1)^2-1+(y-1)^2-1=0$$

Simplify by combining the constants:

$$(x-1)^2+(y-1)^2-2=0 → (x-1)^2+(y-1)^2=2$$

This is the standard equation of a circle centered at \((1, 1)\) with radius \(\sqrt{2}.\) Therefore, this equation represents a real circle.

Analyze the equation \(5x^2-2y^2-10x+8y-13=0.\) It contains quadratic terms in \(x\) and \(y\) with opposite signs (\(5\) and \(-2\)), suggesting a hyperbola. We complete the square to verify.

Group the \(x\) and \(y\) terms:

$$5x^2-10x-2y^2+8y=13$$

Factor out the leading coefficients of \(x^2\) and \(y^2\):

$$5(x^2-2x)-2(y^2-4y)=13$$

Complete the square for \(x\):

$$x^2-2x=(x^2-2x+1)-1=(x-1)^2-1$$

For \(y\):

$$y^2-4y=(y^2-4y+4)-4=(y-2)^2-4$$

Substitute into the equation:

$$5[(x-1)^2-1]-2[(y-2)^2-4]=13$$

Simplify:

$$5(x-1)^2-5-2(y-2)^2+8=13$$

$$5(x-1)^2-2(y-2)^2+3=13$$

$$5(x-1)^2-2(y-2)^2=10$$

Divide the entire equation by 10 to obtain standard form:

$$\dfrac{(x-1)^2}{2}-\dfrac{(y-2)^2}{5}=1$$

This is the standard equation of a hyperbola centered at \((1, 2),\) with a horizontal transverse axis parallel to the x-axis, where \(a^2=2\) and \(b^2=5\).

Examine the equation \(y^2+9x+9=0.\) Notice there is only one squared term (\(y^2\)) and a linear term in \(x,\) which indicates a parabola. Rearrange into standard form:

$$y^2=-9x-9$$

$$y^2=-9(x+1)$$

This represents a horizontal parabola opening to the left, with vertex at \((-1, 0)\) and parameter \(p=-\dfrac{9}{4}\).

Analyze the equation \(x^2+y^2+4x+5=0.\) It contains squared terms in \(x\) and \(y\) with equal coefficients (both 1), suggesting a circle. We complete the square for \(x\) (there is no linear term in \(y\)):

Group terms in \(x\):

$$x^2+4x+y^2=-5$$

Complete the square for \(x\):

$$x^2+4x=(x^2+4x+4)-4=(x+2)^2-4$$

Substitute into the equation:

$$(x+2)^2-4+y^2=-5$$

$$(x+2)^2+y^2=-1$$

This equation states that the sum of two real squares equals \(-1,\) which is impossible in the real number system. Therefore, the equation represents no real geometric locus.

Examine the equation \(16x^2+9y^2-64x-54y+1=0.\) The coefficients of \(x^2\) and \(y^2\) are positive and unequal, which suggests an ellipse. We complete the square:

Group the \(x\) and \(y\) terms:

$$16x^2-64x+9y^2-54y=-1$$

Factor out the coefficients:

$$16(x^2-4x)+9(y^2-6y)=-1$$

Complete the square for \(x\):

$$x^2-4x=(x^2-4x+4)-4=(x-2)^2-4$$

For \(y\):

$$y^2-6y=(y^2-6y+9)-9=(y-3)^2-9$$

Substitute back into the equation:

$$16[(x-2)^2-4]+9[(y-3)^2-9]=-1$$

$$16(x-2)^2-64+9(y-3)^2-81=-1$$

$$16(x-2)^2+9(y-3)^2-145=-1$$

$$16(x-2)^2+9(y-3)^2=144$$

Divide by 144 to obtain standard form:

$$\dfrac{(x-2)^2}{9}+\dfrac{(y-3)^2}{16}=1$$

This is the standard equation of a vertical ellipse centered at \((2, 3),\) with semi-major axis \(a=4\) (parallel to the y-axis) and semi-minor axis \(b=3\).

Analyze the equation \(-7x^2+12y^2+28x+72y-4=0.\) The quadratic terms have coefficients with opposite signs, indicating a hyperbola. We complete the square:

Rearrange the terms:

$$-7x^2+28x+12y^2+72y=4$$

Factor out the coefficients:

$$-7(x^2-4x)+12(y^2+6y)=4$$

Complete the square for \(x\):

$$x^2-4x=(x^2-4x+4)-4=(x-2)^2-4$$

For \(y\):

$$y^2+6y=(y^2+6y+9)-9=(y+3)^2-9$$

Substitute back into the equation:

$$-7[(x-2)^2-4]+12[(y+3)^2-9]=4$$

$$-7(x-2)^2+28+12(y+3)^2-108=4$$

$$-7(x-2)^2+12(y+3)^2-80=4$$

$$-7(x-2)^2+12(y+3)^2=84$$

Divide by 84 and rearrange:

$$\dfrac{(y+3)^2}{7}-\dfrac{(x-2)^2}{12}=1$$

This is the standard equation of a hyperbola centered at \((2, -3),\) with a vertical transverse axis parallel to the y-axis, where \(a^2=7\) and \(b^2=12\).

Analyze the equation \(x^2+y^2-6x-8y+40=0.\) The squared terms have equal coefficients (1 for \(x^2\) and \(y^2\)), suggesting a potential circle. We complete the square for both variables:

Group the \(x\) and \(y\) terms:

$$x^2-6x+y^2-8y=-40$$

Complete the square for \(x\):

$$x^2-6x=(x^2-6x+9)-9=(x-3)^2-9$$

For \(y\):

$$y^2-8y=(y^2-8y+16)-16=(y-4)^2-16$$

Substitute into the equation:

$$(x-3)^2-9+(y-4)^2-16=-40$$

$$(x-3)^2+(y-4)^2-25=-40$$

$$(x-3)^2+(y-4)^2=-15$$

The sum of squares equals a negative number (-15), which has no solution in the real numbers. Therefore, this equation represents no real geometric locus.

Analyze the equation \(x^2+6x-8y+41=0.\) Since there is only one squared term (\(x^2\)) and a linear term in \(y,\) it represents a parabola. Rearrange the terms:

$$x^2+6x=8y-41$$

Complete the square for \(x\):

$$x^2+6x=(x^2+6x+9)-9=(x+3)^2-9$$

Substitute back:

$$(x+3)^2-9=8y-41$$

$$(x+3)^2=8y-32$$

$$(x+3)^2=8(y-4)$$

This is the standard equation of a parabola opening upward, with vertex at \((-3, 4)\) and focal parameter \(p=2\).