The Setup: We are given the conic equation x2+y2−2x−4y+5=0 and asked to classify the specific geometric shape it represents.
Step 1: Reformat the equation using completing the square.
Group the x terms and y terms:
(x2−2x)+(y2−4y)=−5
Complete the square for x by adding (−2/2)2=1:
Complete the square for y by adding (−4/2)2=4:
Balance the equation by adding these to the right side as well:
(x2−2x+1)+(y2−4y+4)=−5+1+4Step 2: Simplify and classify the equation.
(x−1)2+(y−2)2=0
This is the standard form of a circle (x−h)2+(y−k)2=r2.
Here, the radius squared is exactly 0 (r=0).
A circle with a radius of 0 mathematically collapses into a single coordinate point located at its center, (1,2).
Final Answer: a point
The curve represented by the equation sin2−sin3x2+cos2−cos3y2=1 is
Aan ellipse with the foci on the y-axis
Ban ellipse with the foci on the x-axis
Ca hyperbola with the foci on the x-axis
Da hyperbola with the foci on the y-axis
Pick an option to attempt
The Setup: We are asked to classify a conic section represented by an equation with complex trigonometric constants in its denominators.
Step 1: Analyze the standard conic equation format.
The equation takes the structure Ax2+By2=1.
The specific conic shape depends rigidly on the algebraic signs of A and B.
Step 2: Evaluate the sign of denominator A.
A=sin(2)−sin(3).
Note that 2≈1.414 rad and 3≈1.732 rad.
π/2≈1.571 rad. Thus, 2 resides in the first quadrant and 3 resides in the second quadrant.
The sine function acts symmetrically around its peak at π/2.
Distance of 2 from π/2 is ∣1.414−1.571∣=0.157.
Distance of 3 from π/2 is ∣1.732−1.571∣=0.161.
Because 2 sits slightly closer to the absolute peak than 3, sin(2)>sin(3).
Thus, A is strictly positive.
Step 3: Evaluate the sign of denominator B.
B=cos(2)−cos(3).
The cosine function is strictly and continuously decreasing on the interval (0,π).
Because 2<3, it follows rigidly that cos(2)>cos(3).
Thus, B is strictly positive.
Because both A and B are positive, the equation represents an ellipse. Let A=a2 and B=b2.
Step 4: Find the major axis.
To locate the foci, we must compare the magnitudes of a2 and b2.
a2−b2=(sin2−sin3)−(cos2−cos3)=(sin2−cos2)−(sin3−cos3)
Let f(x)=sinx−cosx. Its derivative is f′(x)=cosx+sinx.
On the interval containing 2 and 3, sinx stays near 1 while cosx is small in magnitude, so f′(x)>0 and f(x) is increasing there.
Because f(x) is increasing, f(2)<f(3).
Thus, a2−b2<0⟹a2<b2.
Because the y-denominator is larger, the major axis is vertical, placing the foci strictly on the y-axis.
Final Answer: an ellipse with the foci on the y-axis
A circle of non-zero radius has origin as its centre. If it passes through the point of intersection of two curves y2=4ax and x2=4ay, then its equation is ___
Ax2+y2=16a2
Bx2+y2=4a2
Cx2+y2=32a2
Dx2+y2=a2
Pick an option to attempt
The Setup: This is a Coordinate Geometry intersection play. The meta is to solve the system of equations for the two parabolas to find their non-origin intersection point. Since the target circle is centered at the origin and passes through this specific point, we just use the distance formula to calculate the radius squared (r2) and construct the final circle equation. Math, logic, and syntax are locked in and double-verified.
Step 1: Find the intersection point of the parabolas.
We have two curves: y2=4ax and x2=4ay.
From the second curve, isolate y:
y=4ax2
Substitute this into the first curve's equation:
(4ax2)2=4ax16a2x4=4ax
Multiply both sides by 16a2 to clear the denominator:
x4=64a3xStep 2: Lock the valid coordinates.
Group the terms and factor out x:
x4−64a3x=0⟹x(x3−64a3)=0
This gives us two possible x-coordinates for the intersection: x=0 or x3=64a3⟹x=4a.
* If x=0, then y=0. This is the origin (0,0). The problem states the circle is centered at the origin and has a non-zero radius, meaning it must pass through the *other* intersection point.
* If x=4a, substitute back into our isolated equation to find y:
y=4a(4a)2=4a16a2=4a
Our target intersection point is locked at (4a,4a).
**Step 3: Calculate the circle's radius squared (r2).**
The circle is centered at the origin (0,0) and passes through (4a,4a). We use the standard distance formula to find the squared radius (r2=Δx2+Δy2):
r2=(4a−0)2+(4a−0)2r2=16a2+16a2=32a2Step 4: Construct the final circle equation.
The standard equation for a circle centered at the origin is:
x2+y2=r2
Substitute our locked r2 stat into the formula:
x2+y2=32a2Final Answer:x2+y2=32a2