Classify before calculating: translations, rotations, degeneracies and intersections across conic families
JEE Main · JEE Advanced85 min study guide10 indexed sections
BY THE END, YOU CAN
Classify general second-degree equations using both quadratic structure and remaining terms.
Translate to a center and rotate axes without losing the coordinate transformation.
Derive focus–directrix equations and distinguish ellipse, parabola and hyperbola conditions.
Solve mixed line/conic and conic/conic problems with degree, reality and branch checks.
BEFORE YOU START
Standard circle, ellipse, parabola and hyperbola equations.
Completing squares, determinants and quadratic forms.
Line equations and elementary trigonometric rotations.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
1. A quadratic equation is more general than a familiar standard conic
A general real second-degree equation has form Ax²+Bxy+Cy²+Dx+Ey+F=0, with at least one of A,B,C nonzero. Its quadratic part controls the main directional behavior, while the linear and constant terms determine position and can cause degeneracy or an empty real locus.
The expression B²−4AC is a useful first classifier. A negative value indicates ellipse-type quadratic structure, zero indicates parabola-type structure and a positive value indicates hyperbola-type structure. These labels do not by themselves guarantee a nondegenerate real ellipse, parabola or hyperbola.
For example, x²+y²=1, x²+y²=0 and x²+y²=−1 have the same quadratic part but describe a circle, a single point and no real points. Similarly x²−y²=1 is a hyperbola while x²−y²=0 is a pair of intersecting lines.
The correct workflow is to identify the quadratic structure, remove linear terms where possible, diagonalize any cross term, and then inspect the resulting equation including its constant. Keep the coordinate transformation so that geometric features can be returned to the original axes.
02 / UNDERSTAND & APPLY
2. The cross-term coefficient is split between two matrix entries
The quadratic part is [x y]M[x y]ᵀ with M=[[A,B/2],[B/2,C]]. Its determinant is AC−B²/4, so B²−4AC is −4det M. Using B instead of B/2 in the matrix doubles the cross term and changes the classification.
An orthogonal rotation diagonalizes this real symmetric matrix. The diagonal coefficients are its real eigenvalues. Same-sign nonzero eigenvalues give a definite quadratic part; opposite signs give an indefinite one; one zero eigenvalue gives rank-one structure. The trace A+C and determinant are unchanged by rotation.
Translation does not change the quadratic part, but it changes the linear and constant terms. Multiplying the whole equation by a nonzero constant leaves the same locus, so normalize thoughtfully before comparing signs or identifying standard parameters.
A circle in equal-scale Cartesian coordinates requires A=C≠0 and B=0 before any translation, because rotations preserve equal eigenvalues. An equation with different squared coefficients may be transformed to a unit circle by nonuniform scaling, but that scaling does not preserve the original Euclidean distances and angles.
03 / UNDERSTAND & APPLY
3. A center removes both linear terms when it exists
Substitute x=h+u,y=k+v. The new linear coefficients are 2Ah+Bk+D and Bh+2Ck+E. A center therefore solves 2Ah+Bk+D=0 and Bh+2Ck+E=0.
If 4AC−B²≠0, this linear system has a unique solution. The equation becomes a centered quadratic form plus a constant. Ellipses and hyperbolas in nondegenerate form have such a center. If the determinant is zero, the center system may be inconsistent or have infinitely many solutions in degenerate cases.
A nondegenerate parabola has no center: one remaining linear term along the zero-eigenvalue direction is essential to its shape. Trying to force both linear terms to zero in that case either fails algebraically or mistakenly discards the term that makes the curve a parabola.
After translating, the signs of the quadratic coefficients and the new constant decide whether the locus is real, empty or degenerate. Do not divide by a parameter-dependent constant before checking whether that constant might be zero.
ORIGINAL WORKED EXAMPLE
Same quadratic type, three different real loci
Classify x²+4y²−2x+8y+c=0 according to the real parameter c.
Complete squares: (x−1)²+4(y+1)²=5−c.
For c<5 the right side is positive, so division gives a real nondegenerate ellipse centered at (1,−1).
For c=5 both nonnegative squared terms must vanish, giving the single point (1,−1).
For c>5 the right side is negative, impossible for the nonnegative left side. The quadratic discriminant is negative in all three cases, so it alone cannot distinguish them.
Answer c<5: ellipse; c=5: one point; c>5: empty real locus.
04 / UNDERSTAND & APPLY
4. Rotate the axes, then translate features back
Use x=u cos θ−v sin θ and y=u sin θ+v cos θ. The new xy-type coefficient becomes (C−A)sin2θ+Bcos2θ. Setting it to zero removes the cross term. When A≠C, tan2θ=B/(A−C) is convenient, but its angle choice must be consistent with the full rotation.
If A=C and B≠0, a 45-degree rotation is natural. Define u=(x+y)/√2 and v=(y−x)/√2, so x=(u−v)/√2,y=(u+v)/√2. Writing both directions of the transformation prevents a focus or vertex from being reported in the wrong coordinate system.
Rotation preserves Euclidean lengths and angles because its matrix is orthogonal. A translation and rotation can therefore put a conic into principal-axis form without changing its actual semi-axis lengths or focal distances. A later nonuniform scaling is a different operation and must not be treated as a rigid motion.
When the rotated equation has one zero quadratic coefficient, inspect the remaining linear term in that direction. A nonzero term gives a parabola. If it also vanishes, the equation depends on only one coordinate and can represent two parallel lines, one repeated line or no real lines.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
The coefficient matrix is [[5,−3],[−3,5]], with eigenvalues 2 and 8.
ORIGINAL WORKED EXAMPLE
Find the original-coordinate foci of a rotated ellipse
Classify 3x²+2xy+3y²=8 and find its foci.
Set u=(x+y)/√2 and v=(y−x)/√2. Then 3x²+2xy+3y²=4u²+2v².
The equation is u²/2+v²/4=1. Its major semi-axis is 2 along the v-axis, minor semi-axis √2 along the u-axis, and focal distance c=√(4−2)=√2.
The foci in rotated coordinates are (u,v)=(0,±√2).
Return using x=(u−v)/√2,y=(u+v)/√2. The original-coordinate foci are (−1,1) and (1,−1). The major vertices are (−√2,√2) and (√2,−√2).
Answer A rotated ellipse with foci (−1,1),(1,−1).
ORIGINAL WORKED EXAMPLE
A rank-one quadratic becomes a rotated parabola
Find focus and directrix of x²+2xy+y²−4x+4y=0.
With the same rotation, (x+y)²=2u² and y−x=√2v. The equation becomes 2u²+4√2v=0, or u²=−2√2v.
This is u²=4pv with p=−√2/2. Its vertex is the origin, focus in rotated coordinates (0,−√2/2), and directrix v=√2/2.
Transform the focus back: x=1/2,y=−1/2.
The directrix equation (y−x)/√2=√2/2 becomes y−x=1. The focus lies on the opposite side of the vertex from this line, as required.
After rotation and translation, a definite quadratic part can yield an ellipse, a point or an empty set according to the remaining constant. An indefinite quadratic part yields a hyperbola when the centered constant is nonzero and two intersecting real lines when it is zero.
For a rank-one quadratic part, a nonzero linear coefficient in the missing-square direction gives a parabola. Otherwise one obtains an equation such as λu²=k: two parallel real lines if k/λ>0, a repeated line if k=0, or an empty set if k/λ<0.
The determinant of the augmented symmetric matrix [[A,B/2,D/2],[B/2,C,E/2],[D/2,E/2,F]] is zero for a degenerate projective conic. In elementary real-coordinate work, completing squares still reveals whether that degeneracy appears as real lines, a point or another real-locus outcome. The matrix test should supplement, not replace, that inspection.
If all quadratic coefficients vanish for a special parameter value, the equation is no longer second degree. It may reduce to a line, a contradiction or the entire plane. Separate such a value before invoking any conic-specific classification.
ORIGINAL WORKED EXAMPLE
A parameter changes both type and degeneracy
Classify x²+λy²=1 for every real λ.
For λ>0, the equation is x²/1+y²/(1/λ)=1, a real ellipse. At λ=1 it is the unit circle.
For λ=0, the equation is x²=1, giving the two vertical parallel lines x=1 and x=−1.
For λ<0, rewrite x²−y²/(−1/λ)=1, a horizontal hyperbola.
The boundary λ=0 has B²−4AC=0 but is not a parabola because no linear y term remains to create one.
Answer λ>0: ellipse (circle at 1); λ=0: x=±1; λ<0: hyperbola.
06 / UNDERSTAND & APPLY
6. Eccentricity unifies three noncircular families
For a fixed focus F and directrix line ℓ not containing F, impose PF=e·distance(P,ℓ) with e>0. A ratio 0<e<1 gives an ellipse, e=1 a parabola and e>1 a hyperbola, with the usual nondegenerate geometric setup.
Take F=(a,0), ℓ:x=−a with a>0. The equation is √[(x−a)²+y²]=e|x+a|. Squaring is reversible because both sides are nonnegative, giving (1−e²)x²−2a(1+e²)x+y²+a²(1−e²)=0.
The sign of 1−e² now explains the classification. For e<1 both squared directions have the same sign after centering; for e>1 they have opposite signs; for e=1 the x² term vanishes and y²=4ax remains.
A circle is not obtained by simply setting e=0 with this fixed finite focus/directrix pair: that would force PF=0 and give only the focus. A circle is an ellipse with coincident foci or an appropriate limiting focus–directrix description with the directrix receding, not a nondegenerate e=0 instance of this fixed setup.
ORIGINAL WORKED EXAMPLE
Build a conic from the distance ratio rather than guessing its center
Find the standard equation of points whose distance to (2,0) is half their distance to x=−2.
The distance equation is √[(x−2)²+y²]=(1/2)|x+2|. Squaring and multiplying by 4 gives 4(x−2)²+4y²=(x+2)².
Expand to get 3x²+4y²−20x+12=0.
Complete the square: 3(x−10/3)²+4y²=64/3. Thus (x−10/3)²/(64/9)+y²/(16/3)=1.
The center is (10/3,0), not the original focus or the origin. Its semi-major axis is 8/3 and focal distance is 4/3, giving eccentricity 1/2 as required; one focus is x=10/3−4/3=2.
Answer (x−10/3)²/(64/9)+y²/(16/3)=1.
07 / UNDERSTAND & APPLY
7. A line substitution must retain its actual degree
Substituting a line parametrization into a conic gives a polynomial of degree at most two in the line parameter. When it is genuinely quadratic, discriminant sign gives two, one repeated or no real intersections. A repeated root at a smooth point corresponds to tangency.
If the quadratic coefficient vanishes, the result can be linear or constant. A line parallel to a hyperbola asymptote may have exactly one finite transverse intersection. Therefore one real intersection by itself is not a universal tangent criterion for all conics.
For F(x,y)=0 at a point P with nonzero gradient, the tangent is F_x(P)(x−x₀)+F_y(P)(y−y₀)=0. This unifies the point-tangent equations for all four standard conics. If the gradient vanishes, the point is singular and a unique smooth tangent cannot be inferred from that formula.
For a pair of intersecting lines, the intersection is singular and has two line directions. For an isolated point conic, there is no ordinary curve tangent there. These degenerate cases are why a smoothness condition belongs with the gradient tangent rule.
ORIGINAL WORKED EXAMPLE
Two intersections merge at a parameter value
For the ellipse x²/4+y²=1 and lines y=m x+2, find the slopes that give tangency.
Substitute the line: x²/4+(mx+2)²=1. The x² coefficient m²+1/4 is always positive, so the equation is genuinely quadratic for every real m.
The quadratic is (m²+1/4)x²+4mx+3=0. Its discriminant is 16m²−12(m²+1/4)=4m²−3.
Tangency requires 4m²−3=0, giving m=±√3/2.
The same answer follows from the ellipse tangent intercept condition 2²=4m²+1. There is no lost degree case here because the leading coefficient cannot vanish.
Answer Tangency slopes m=±√3/2.
08 / UNDERSTAND & APPLY
8. Eliminate one structure, then verify both originals
For two conics, subtraction can cancel common quadratic terms and reduce the intersection problem to a line plus one conic. This is the radical-axis method for two normalized circles, but the same algebraic idea can work for other pairs with shared terms.
When equations involve x² and y² only, solve first for X=x² and Y=y². The resulting linear system may be easy, but real coordinates require X,Y≥0. Positive values generate independent sign choices; zero values reduce the number of distinct points.
Elimination can create candidates that fail a denominator restriction or a branch condition in the original geometric description. Substitute every final point into both equations. If a locus was defined by a signed distance difference, the squared conic equation may contain an unwanted branch.
Two distinct conics without a shared component have at most four complex projective intersection points counted with multiplicity; in elementary problems, one usually solves the real candidates directly. If they share a line component, there can be infinitely many common points, so a blanket “at most four real points” statement needs that no-common-component condition.
ORIGINAL WORKED EXAMPLE
Use squared coordinates and keep all real signs
Find the intersections of x²+y²=13 and x²/9+y²/4=1.
Let X=x²,Y=y². The equations become X+Y=13 and 4X+9Y=36.
Substituting X=13−Y gives 52+5Y=36, so Y=−16/5.
A real square cannot be negative. Therefore there are no real intersections, without needing to solve for x or y.
This matches a geometric bound: every point on the ellipse has x²+y²≤9, while the circle requires 13.
Answer There are no real intersection points.
09 / UNDERSTAND & APPLY
9. Original practice: classification and mixed reasoning
For every classification, state the real-locus outcome as well as the sign of the quadratic discriminant. For a transformed feature, return to original coordinates.
ORIGINAL PRACTICE · 1
Classify x²+y²+1=0 over the reals.
Reveal answer & reasoning
Empty real locus.
The left side is at least 1. Its ellipse-type quadratic part does not guarantee a real ellipse.
ORIGINAL PRACTICE · 2
Classify x²−4y²=0.
Reveal answer & reasoning
Two intersecting lines x=±2y.
Factor as (x−2y)(x+2y)=0; the centered constant is zero.
ORIGINAL PRACTICE · 3
Classify x²−4x−y+7=0.
Reveal answer & reasoning
A parabola: y=(x−2)²+3.
Complete the square to identify vertex (2,3), with a nonzero linear y term.
ORIGINAL PRACTICE · 4
Find the center of 2x²+3y²−8x+6y−1=0.
Reveal answer & reasoning
(2,−1).
The center equations are 4h−8=0 and 6k+6=0. Completing squares gives the same point.
ORIGINAL PRACTICE · 5
For 5x²+6xy+5y²=8, find the diagonal coefficients under u=(x+y)/√2,v=(y−x)/√2.
Reveal answer & reasoning
8u²+2v²=8.
The symmetric matrix has eigenvalues 8 and 2 along the stated u,v directions.
ORIGINAL PRACTICE · 6
Classify y²=−4 over the real plane.
Reveal answer & reasoning
Empty real locus.
No real y satisfies it. The absence of x does not make it a parabola.
ORIGINAL PRACTICE · 7
Classify (y−2)²=0.
Reveal answer & reasoning
The repeated line y=2.
Every x is allowed but y is fixed. It is a degenerate second-degree equation.
ORIGINAL PRACTICE · 8
Which family has focus–directrix ratio e=3/2?
Reveal answer & reasoning
A hyperbola, for the nondegenerate focus/directrix setup.
The eccentricity exceeds 1, making the centered quadratic structure indefinite.
ORIGINAL PRACTICE · 9
Find intersections of x²+y²=5 and x²−y²=3.
Reveal answer & reasoning
(±2,±1), all four independent sign choices.
Adding gives x²=4 and subtracting gives y²=1. Each sign combination satisfies both equations.
ORIGINAL PRACTICE · 10
Does x²−y²=1 meet the line y=x?
Reveal answer & reasoning
No real or complex finite intersection.
Substitution makes the left side zero, so the equation becomes 0=1. The line is an asymptote.
ORIGINAL PRACTICE · 11
Find the tangent to x²+y²=5 at (1,2) using the gradient.
Reveal answer & reasoning
x+2y=5.
The gradient is (2,4), so 2(x−1)+4(y−2)=0.
ORIGINAL PRACTICE · 12
Can the gradient tangent formula give a unique tangent to x²−y²=0 at the origin?
Reveal answer & reasoning
No.
The gradient is zero there and the locus consists of two intersecting lines y=±x, so there are two branch directions.
10 / UNDERSTAND & APPLY
10. Classification is a sequence of checks
First check that the equation is genuinely second degree. Next inspect the quadratic matrix, then translate and rotate as appropriate. Finally inspect the remaining constant and linear terms to distinguish a real conic from lines, a point or an empty set.
Preserve the transformations and their geometric meaning. Rotation and translation preserve Euclidean geometry; nonuniform scaling preserves affine relations but not general distances or angles. Report foci, vertices and directrices in the coordinate system actually requested.
In intersection problems, retain reality conditions on squared coordinates and check the actual polynomial degree after substitution. At a tangent contact, verify that the point lies on the conic and the gradient is nonzero before using the smooth tangent formula.
The unifying idea is that the same quadratic algebra can encode very different real geometry. Discriminants, determinants and standard forms are tools for uncovering that geometry, not substitutes for checking the hypotheses and exceptional cases.
KEEP THE BIG IDEAS
Quick revision
The quadratic discriminant identifies a type of structure, not the complete real locus.
Translation removes linear terms at a center; rotation removes a cross term.
Degenerate and empty cases must be checked after normalization.
Focus–directrix eccentricity separates ellipse, parabola and hyperbola for e>0.
Line and conic intersections require degree, reality and smoothness checks.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.