Construct conic loci and convert coordinate forms

Coordinate descriptions need restrictions

Describe a locus using distances, Cartesian equations, parameters or polar coordinates. When converting, retain parameter intervals, nonnegative radii and sign conditions so that the new description represents the intended set of points

A locus is a set of points satisfying a condition

A line can be defined by Ax+By+C=0 with(A,B) nonzero,or as the perpendicular bisector of two distinct points. A circle is the set at a fixed positive distance from a centre. A parameter or polar relation must specify its allowed values; the equation alone may describe a larger set

A direction vector gives a parametric line

Eliminate the parameter from the coordinate rules. The unrestricted parameter traces the whole line; the closed interval traces a segment between the stated endpoints, and nonnegative values trace a ray. A vertical line has a horizontal direction component of zero

Parametric segment

\(\displaystyle x=1+2t,\quad y=-2+t,\quad -1\le t\le2\)

For the closed segment above,give the endpoint x-coordinate at t=2.

A parabola compares a point distance with a line distance

A parabola is the set equidistant from a focus and a directrix. For(y−k)²=4p(x−h),the vertex is(h,k),focus(h+p,k) and directrixx=h−p,where p≠0. Interchange x and y for a vertical axis. A negative p reverses the opening direction

In the shown parabola,why doesP belong to the locus?

Recover a parabola from its construction data

For focus(2,−1) and directrixx=−2,the vertex is(0,−1),p=2 and(y+1)²=8x. The point(2,3) is distance4 from the focus and distance4 from the directrix. The focus itself is not on the parabola

Focus

\(\displaystyle (2,-1)\)

Directrix

\(\displaystyle x=-2\)

Give the exact value of 4p

Ellipse and hyperbola use different focal conditions

An ellipse has distance sum2a with a>c>0; then b²=a²−c². A hyperbola has absolute distance difference2a with c>a>0; then b²=c²−a². Thus c²=a²−b² for the ellipse but c²=a²+b² for the hyperbola. The equality cases need separate treatment,as in9C

Read the hyperbola’s positive term first

Read the positive squared term to identify the transverse axis. The conjugate-axis construction points are not points on the hyperbola

The centre is (1, −2), the transverse semiaxis is 3 and the conjugate scale is 4

Hyperbola

\(\displaystyle \frac{(x-1)^2}{9}-\frac{(y+2)^2}{16}=1\)

Give c² for this hyperbola

Hyperbola

\(\displaystyle \frac{(x-1)^2}{9}-\frac{(y+2)^2}{16}=1\)

What does the Cartesian hyperbola equation describe?

Circle and ellipse parametrisations retain an interval

For x=h+a cos t,y=k+b sin t with a,b>0,use cos²t+sin²t=1 to obtain(x−h)²/a²+(y−k)²/b²=1. Equal a,b give a circle. The interval0≤t<2π traces the entire curve once;0≤t≤π traces only its upper half. Coordinate bounds describe a box,not the locus by themselves

Which part is traced by x=1+4cos t,y=−2+3sin t,0≤t≤π?

Arc

\(\displaystyle x=1+4\cos t,\quad y=-2+3\sin t,\quad 0\le t\le\pi\)

Give the maximum y-coordinate on that arc

Eliminating a parameter can lose the traced range

For x=1+2t,y=−1+t²,−1≤t≤2,the Cartesian equation is(x−1)²=4(y+1). The actual arc has−1≤x≤5 and−1≤y≤3. Its endpoints are(−1,0),(5,3),but its lowest point(1,−1) occurs at t=0,not at either parameter endpoint

Restricted arc

\(\displaystyle x=1+2t,\quad y=-1+t^2,\quad -1\le t\le2\)

Give the minimum y-coordinate of this restricted arc

Secant and tangent can trace both hyperbola branches

The parametrisation x=1+3sec t,y=−2+4tan t gives(x−1)²/9−(y+2)²/16=1. Use0≤t<2π withπ/2 and3π/2 excluded. Positive secant gives the right branch and negative secant gives the left. At t=π/3,the point is(7,−2+4√3)

Horizontal coordinate

\(\displaystyle x=1+3\sec t\)

Give the x-coordinate at t=π/3 in this parametrisation

State the polar convention before converting

Here r≥0,x=r cosθ and y=r sinθ;angles are in radians. A nonzero point has angle values differing by2π. The origin has no unique direction. Substitute x,y and r²=x²+y²,but retain denominator and sign conditions. A Cartesian identity alone does not prove the polar chart covers every Cartesian point

A polar line has an angular restriction

The relation r cosθ=3 gives x=3. Written as r=3/cosθ with r≥0,it requires cosθ>0;zero gives no point and negative cosine would require a negative r. The Cartesian line contains every y value despite that angular restriction

A polar circle needs an origin check

For r=4cosθ,multiply by r to obtain x²+y²=4x,or(x−2)²+y²=4. This circle has centre(2,0),radius2 and includes the origin. With r≥0,retain cosθ≥0. At cosθ=0,the original polar relation gives r=0,so multiplication did not add a false origin

Polar circle

\(\displaystyle r=4\cos\theta,\quad r\ge0\)

Give the squared radius of that circle

Focus–directrix form connects the three conics

For focus at the origin and directrixx=d/e with d>0,e>0,the distance condition is r=e|x−d/e|. On the side x≤d/e,this becomes r=d−ex and hence r=d/(1+e cosθ). It is an ellipse for0<e<1,a parabola for e=1 and a hyperbola branch for e>1. Always require the denominator positive

Keep the unsquared condition

Squaring r=d−ex gives(1−e²)x²+y²+2dex−d²=0,together with d−ex≥0. The last inequality is essential: it states that the right side was a nonnegative distance. A sine denominator interchanges the axes; a minus sign reverses the associated directrix direction

Worked polar ellipse and parabola

For r=3/(1+cosθ/2),the Cartesian relation is3x²/4+y²+3x−9=0,or(x+2)²/16+y²/12=1. Its focus is the origin and its directrix isx=6. For r=2/(1+cosθ),the relation is y²=−4(x−1),a left-opening parabola with vertex(1,0),focus(0,0) and directrixx=2.

Polar ellipse

\(\displaystyle r=\frac{3}{1+\frac12\cos\theta},\quad r\ge0\)

Give the centre’s x-coordinate for the worked polar ellipse

Squaring a polar hyperbola can add a branch

For r=3/(1+2cosθ),the squared equation is(x−2)²−y²/3=1. But3−2x≥0 must remain. It selects the left branch x≤1 and excludes the right branch x≥3. The point(3,0) solves the squared equation but would require r=−3 in the unsquared one

Polar relation

\(\displaystyle r=\frac{3}{1+2\cos\theta},\quad r\ge0\)

Which branch is represented by that nonnegative-radius polar formula?

Construct the loci from distances,ungraded

Construct a line as a perpendicular bisector,a circle from centre and radius,a focus–directrix parabola,an ellipse from a fixed distance sum and a hyperbola from a fixed absolute distance difference. Explain the nondegenerate parameter conditions and label the exact defining data. These constructions are ungraded

Model: use the distance definition for each construction

Equal distances to(−2,0),(2,0) give the linex=0. A compass centred atC with positive radiusR gives its circle. For the parabola,take lines parallel to the directrix at distanceq and intersect them with the circle centred atF of radiusq. For an ellipse,choose distancesu,v with u+v=2a>2c and intersect circles about its two foci; a taut string of length2a gives the same set. For a hyperbola,use|u−v|=2a<2c and circle intersections on both sides. Check the defining distances; reference foci and construction-box corners are not automatically locus points

Prove the conversion in both directions,ungraded

Derive(y−k)²=4p(x−h) from focus(h+p,k) and directrixx=h−p,including p<0. Then prove the converse and explain why the same unchecked squaring can fail for a polar hyperbola. This written proof is ungraded

Model: distances and sign conditions control the converse

The focus distance squared is(x−h−p)²+(y−k)²; the line distance squared is(x−h+p)². Equating and expanding gives(y−k)²=4p(x−h). Conversely,this equation makes those nonnegative squared distances equal,so the distances are equal. No additional sign restriction is needed because both original quantities were distances. In r=d−ex,the right side is not automatically nonnegative,so after squaring it must still satisfy d−ex≥0.

Compare coordinate forms with technology,ungraded

Plot the full Cartesian ellipse and its upper-half parametrisation, then the Cartesian hyperbola and its positive-radius polar chart. Record the parameter intervals and missing points or branches. Explain why matching a few plotted points does not establish equal loci. This technology task is ungraded

Model: compare sets,not only equations

The ellipse parametrisation obeys its Cartesian equation but adds a restriction that omits the lower half. The polar hyperbola obeys its Cartesian equation but retains a restriction that omits the right branch. Sample interior points, boundaries and excluded angles; use the exact conditions to decide coverage. Continuous curves on a screen are approximations, and a plotting window may hide an unbounded branch