Equal distances to a focus and a line: parameter points, tangents, normals and focal chords
JEE Main · JEE Advanced85 min study guide12 indexed sections
BY THE END, YOU CAN
Derive a parabola from the focus–directrix definition and orient translated forms.
Use the parameter (at²,2at) to describe contacts, tangents and chords.
Find normals through a point without assuming uniqueness.
Connect focal chords, perpendicular tangents and the directrix through exact algebra.
BEFORE YOU START
Cartesian distance, line equations and completing squares.
Quadratic and cubic equations, discriminants and derivatives.
Basic tangent and perpendicularity concepts.
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. The focus and directrix define the curve
A parabola is the locus of points equally distant from a fixed point, the focus, and a fixed line, the directrix, with the focus not on that line. The axis passes through the focus perpendicular to the directrix. The vertex lies halfway between the focus and its perpendicular foot on the directrix.
For focus F=(a,0) and directrix x=−a with a>0, a point P=(x,y) satisfies √[(x−a)²+y²]=|x+a|. Squaring gives y²=4ax. This equation forces x≥0, so x+a>0 and the squared equation indeed recovers the original nonnegative distance equality.
The standard parabola y²=4ax opens right, has vertex (0,0), focus (a,0), axis y=0 and directrix x=−a. Its eccentricity is 1 because the focus-distance to directrix-distance ratio is exactly 1. The parameter a is the vertex-to-focus distance, not the latus rectum length.
A signed parameter p can describe orientation: (y−k)²=4p(x−h), p≠0, has focus (h+p,k) and directrix x=h−p. A positive p opens right and a negative p opens left. The vertical form (x−h)²=4p(y−k) is interpreted by exchanging the coordinate roles.
02 / UNDERSTAND & APPLY
2. Completing one square reveals the vertex
A parabola in an axis-aligned coordinate system has one squared variable and one linear variable after simplification. Complete the square in the squared variable, then isolate the other. The coefficient must be compared with 4p, not with p directly.
The latus rectum is the focal chord perpendicular to the axis. For y²=4ax, set x=a to get y=±2a. Its endpoints are (a,±2a) and its length is 4a. Under a signed orientation parameter p, the geometric length is 4|p|.
The axis is a symmetry line, but the parabola has no center of symmetry. Reflecting a right-opening parabola through a finite point would reverse its unbounded opening direction, so it cannot map the whole curve to itself. This distinguishes it from ellipses and hyperbolas.
An arbitrary quadratic polynomial with only one squared variable need not describe a nondegenerate parabola if the other linear coefficient vanishes. It may instead represent parallel lines, a repeated line or an empty set. Check that the completed equation really has a nonzero p.
ORIGINAL WORKED EXAMPLE
Read a translated left-opening parabola
Find vertex, focus, directrix and latus rectum length for y²−4y+8x+12=0.
Complete the square: (y−2)²+8x+8=0, so (y−2)²=−8(x+1).
This is (y−k)²=4p(x−h) with h=−1,k=2,p=−2.
The vertex is (−1,2), the focus is (h+p,k)=(−3,2), and the directrix is x=h−p=1.
The parabola opens left and its latus rectum length is 4|p|=8. The focus lies to the left of the vertex, consistent with the opening direction.
For y²=4ax with a>0, set y=2at. Then x=at², giving P(t)=(at²,2at) for every real t. Conversely each curve point has the unique parameter t=y/(2a), so this parametrization covers the whole parabola exactly once.
The parameter sign identifies the upper or lower half, while t=0 gives the vertex. Opposite parameters t and −t give reflected points across the axis. Unlike an ellipse parameter, t is not an angle and is not periodic.
The distance from P(t) to the focus is a(1+t²). Indeed, the squared distance is a²(t²−1)²+4a²t²=a²(t²+1)², whose positive square root is a(1+t²). The distance to x=−a is x+a=a(t²+1), independently confirming the locus definition.
For a translated right-opening parabola, add the vertex coordinates: P(t)=(h+at²,k+2at). If the parabola opens in another direction, transform the coordinate roles or use the signed standard form consistently rather than silently retaining the right-opening formulas.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Parameter t=2, a=1 gives P=(at²,2at)=(4,4).
ORIGINAL WORKED EXAMPLE
A parameter point verifies the distance definition
For y²=8x, use t=3/2 to find a point and its distance to the focus.
Here 4a=8, so a=2. The parameter point is (2·9/4,4·3/2)=(9/2,6).
The focus is (2,0). The distance is √[(9/2−2)²+6²]=√(25/4+36)=13/2.
The parameter formula gives a(1+t²)=2(1+9/4)=13/2, the same result.
The directrix is x=−2, and the point-to-line distance is 9/2+2=13/2. All three calculations agree.
Answer P=(9/2,6); focus distance and directrix distance both 13/2.
04 / UNDERSTAND & APPLY
4. The parameter tangent includes the vertical vertex tangent
Implicit differentiation of y²=4ax gives 2yy′=4a, so y′=2a/y when y≠0. At P(t) with t≠0, the tangent slope is 1/t. Writing the point-slope equation and simplifying gives ty=x+at².
The parameter equation remains valid at t=0, where it becomes x=0, the vertical tangent at the vertex. This is a useful example of an implicit line formula extending through a case where the slope formula divides by zero.
At a general on-curve point (x₁,y₁), the tangent is yy₁=2a(x+x₁). The condition y₁²=4ax₁ is essential. Substituting an arbitrary off-curve point into this expression does not give a tangent at that point.
The tangent has a direct supporting-line proof. At another parameter s, x+at²−ty=a(s²+t²−2st)=a(s−t)²≥0, with equality only when s=t. Thus every parabola point lies on one side and the tangent touches at a unique parameter.
ORIGINAL WORKED EXAMPLE
A tangent from a rational contact parameter
Find the tangent to y²=8x at P=(9/2,6).
The parameter is t=y/(2a)=6/4=3/2, with a=2.
The tangent is (3/2)y=x+2·(9/4)=x+9/2, or 2x−3y+9=0.
Substituting P gives 9−18+9=0. Its slope is 2/3, matching 1/t.
The supporting-line residual at parameter s is 2(s−3/2)², which is nonnegative and vanishes only at the contact.
Answer Tangent 2x−3y+9=0.
05 / UNDERSTAND & APPLY
5. Contact parameters turn an exterior-tangent problem into a quadratic
A tangent at parameter t passes through Q=(h,k) exactly when tk=h+at², or at²−kt+h=0. Each distinct real root gives one tangent contact. The discriminant is k²−4ah, so there are two, one or no real tangents according as it is positive, zero or negative.
The zero-discriminant case is precisely Q on the parabola. The region h>k²/(4a), to the right of the curve at the same height, has no real tangents. The other side has two, with the vertex tangent x=0 included automatically when h=0 gives a root t=0.
A finite-slope tangent can be written y=mx+a/m with m≠0, obtained by setting m=1/t. The vertex tangent is missing from that form and must be added separately. A horizontal finite tangent does not exist on this right-opening parabola.
If two distinct tangent parameters are t₁,t₂, subtract their equations to find their intersection: y=a(t₁+t₂), x=at₁t₂. These formulas are useful for reverse problems involving a tangent pair, but assume t₁≠t₂ so that the intersection is not an entire repeated line.
ORIGINAL WORKED EXAMPLE
Find two tangents and verify both contact points
Find tangents from Q=(−2,3) to y²=4x.
Here a=1. The contact-parameter equation is t²−3t−2=0, with roots t₁=(3+√17)/2 and t₂=(3−√17)/2.
The tangents are t₁y=x+t₁² and t₂y=x+t₂². Each has contact P(tᵢ)=(tᵢ²,2tᵢ).
The roots satisfy t₁+t₂=3 and t₁t₂=−2. Their tangent intersection is therefore (−2,3), as requested.
The discriminant is 17>0, confirming two distinct real contacts. Keeping the exact parameter form is cleaner than expanding two radical line equations.
Answer Tangents ty=x+t² for t=(3±√17)/2, touching at (t²,2t).
06 / UNDERSTAND & APPLY
6. A point can lie on several normals
At parameter t, the tangent line ty=x+at² has direction (t,1). A perpendicular normal direction is (1,−t), since their dot product is t−t=0. This vector works for all t, including t=0.
The normal through P(t) therefore has equation y−2at=−t(x−at²), or y=−tx+2at+at³. At t=0 it becomes y=0, the axis. A slope-based calculation also gives −t for nonzero t, but the final formula includes the vertex.
A normal through Q=(h,k) requires at³+(2a−h)t−k=0. This cubic may have one or three distinct real roots, or repeated roots leading to fewer distinct contacts. A single given point does not guarantee a unique normal.
Always verify each real parameter by its contact point and normal equation. A repeated algebraic root does not create an extra geometric line, and complex parameter roots do not represent real contact points.
ORIGINAL WORKED EXAMPLE
Three distinct normals through one point
Find all normals from Q=(3,0) to y²=4x.
With a=1,h=3,k=0, the normal parameter equation is t³−t=t(t−1)(t+1)=0.
The real parameters are 0,1,−1, giving contacts (0,0),(1,2),(1,−2).
The normal equations are y=0 for t=0, y=−x+3 for t=1, and y=x−3 for t=−1.
All three pass through (3,0), and their tangent slopes at the nonvertex contacts are 1 and −1, respectively, perpendicular to the corresponding normal slopes.
Answer Normals y=0, y=−x+3 and y=x−3.
07 / UNDERSTAND & APPLY
7. Two parameters give a chord without solving intersections again
For distinct parameters t₁,t₂, the chord joining P(t₁),P(t₂) has equation (t₁+t₂)y=2x+2at₁t₂. Substituting either parameter verifies it. It can also be derived from the two-point determinant equation without dividing by t₁+t₂.
If t₁+t₂=0, the chord is vertical and the same formula gives x=−at₁t₂=at₁². Thus the implicit expression includes a case that a finite chord-slope formula would miss.
A genuine two-endpoint focal chord passes through F=(a,0). Substitution into the chord equation gives t₁t₂=−1. Both parameters are therefore nonzero and of opposite signs. The axis itself passes through the focus but meets the parabola at only the vertex as a finite point, so it is not a two-endpoint focal chord of this kind.
The chord of contact of tangents from an exterior Q=(h,k) is ky=2a(x+h). Its parameters satisfy sum k/a and product h/a from the tangent quadratic, so the general chord formula immediately gives this equation. It joins contacts and is distinct from the tangent lines through Q.
ORIGINAL WORKED EXAMPLE
A focal chord and perpendicular endpoint tangents
For y²=4x, a focal chord has one endpoint at parameter t₁=2. Find the other endpoint, chord and tangent intersection.
The focal condition gives t₂=−1/2. The endpoints are (4,4) and (1/4,−1).
The chord equation is (3/2)y=2x−2, or 4x−3y−4=0. It contains the focus (1,0).
The tangent intersection is (t₁t₂,t₁+t₂)=(−1,3/2), lying on the directrix x=−1.
The tangent slopes are 1/2 and −2, whose product is −1. Thus the endpoint tangents are perpendicular.
Answer Other endpoint (1/4,−1); chord 4x−3y−4=0; tangent intersection (−1,3/2).
08 / UNDERSTAND & APPLY
8. The directrix is also a locus of tangent intersections
For two distinct nonvertex contacts, tangent slopes are 1/t₁ and 1/t₂. They are perpendicular when t₁t₂=−1. Their intersection has x-coordinate at₁t₂=−a, so it lies on the directrix.
Conversely, for any finite point Q=(−a,k) on the directrix, the tangent-parameter equation is at²−kt−a=0. Its discriminant k²+4a² is positive, so there are two distinct real contacts, and their product is −1. Their tangents are therefore perpendicular.
This establishes the entire directrix as the locus of intersections of perpendicular tangent pairs. The corresponding contact chord is focal, because the same product condition is exactly t₁t₂=−1. A single algebraic relation links the focus and directrix constructions.
The statement concerns tangent pairs, not normals. Normals through a point satisfy a cubic and have a different geometry. Keeping the two parameter equations separate prevents a tangent-locus result from being misapplied to normal problems.
09 / UNDERSTAND & APPLY
9. Parallel chords and reflected directions
The midpoint of a parameter chord is M=(a(t₁²+t₂²)/2,a(t₁+t₂)). If the chord direction is fixed, its slope 2/(t₁+t₂) is fixed when the denominator is nonzero. Hence the y-coordinate of the midpoint is fixed, so midpoints of parallel nonvertical chords lie on a line parallel to the axis.
For a vertical chord, t₂=−t₁ and its midpoint is on y=0, again parallel to the axis. These midpoint loci are called diameters of the parabola, but unlike central-conic diameters they do not all pass through a common center.
The equal-distance definition also produces the reflection property. At P, let D be the perpendicular foot on the directrix. The tangent bisects the angle between the focal segment and the axis-parallel direction in the appropriate internal or external sense. One can verify this through the tangent direction and the unit vector toward the focus.
For P(t), the unit vector from P toward F is ((1−t²)/(1+t²),−2t/(1+t²)). Reflecting the axis-parallel incoming direction (−1,0) across the tangent direction (t,1) yields exactly that vector. Thus rays parallel to the axis and traveling toward the opening’s vertex direction reflect toward the focus under the ideal equal-angle law.
10 / UNDERSTAND & APPLY
10. The latus rectum is the shortest focal chord
For a focal chord, the parameters are t and −1/t with t≠0. The focus lies between its endpoints: their y-coordinates have opposite signs, the connecting line passes through the focus on y=0, and a straight segment crosses that axis between them.
The chord length is therefore the sum of the two focus distances, a(1+t²)+a(1+1/t²)=a(2+t²+1/t²). This is a sum of ordinary nonnegative lengths, not a directed-distance subtraction.
Since (t−1/t)²≥0, one has t²+1/t²≥2, giving chord length at least 4a. Equality requires t=±1, which yields the endpoints (a,±2a) of the latus rectum. Thus the perpendicular focal chord is the shortest focal chord.
The bound supplies a useful feasibility check. A proposed focal chord shorter than 4a cannot exist. There is no finite upper bound: as t approaches zero, the other parameter diverges and the chord grows without bound. The excluded axis direction appears as a limiting infinite-length configuration rather than an ordinary finite chord.
ORIGINAL WORKED EXAMPLE
Check a focal chord length against its minimum
For y²=8x, a focal chord has parameters 2 and −1/2. Find its length.
The scale is a=2. The focus distances are 2(1+4)=10 and 2(1+1/4)=5/2.
The focus lies between the endpoints, so the chord length is 10+5/2=25/2.
The endpoints are (8,8) and (1/2,−2). Their displacement is (15/2,10), whose length is √(225/4+100)=25/2.
The shortest possible focal chord has length 4a=8. The computed 25/2 exceeds that bound, as expected for a nonvertical focal chord.
Answer Chord length 25/2; the minimum focal-chord length is 8.
11 / UNDERSTAND & APPLY
11. Original practice with full solutions
Unless otherwise stated, use y²=4ax with a>0. Preserve vertical tangents and count distinct real normal parameters.
ORIGINAL PRACTICE · 1
Find focus and directrix of y²=12x.
Reveal answer & reasoning
Focus (3,0); directrix x=−3.
Here 4a=12, so a=3.
ORIGINAL PRACTICE · 2
Find vertex and focus of (x−2)²=−8(y+1).
Reveal answer & reasoning
Vertex (2,−1); focus (2,−3).
This vertical form has p=−2, so the focus lies 2 units below the vertex.
ORIGINAL PRACTICE · 3
Find the latus rectum length of y²=20x.
Reveal answer & reasoning
20.
The parameter is a=5 and latus rectum length is 4a.
ORIGINAL PRACTICE · 4
For y²=4x, find the point with parameter t=−3.
Reveal answer & reasoning
(9,−6).
Use (t²,2t) for a=1.
ORIGINAL PRACTICE · 5
Find the tangent to y²=4x at the vertex.
Reveal answer & reasoning
x=0.
The parameter t=0 gives ty=x+t², hence x=0; the slope formula 1/t is unavailable.
ORIGINAL PRACTICE · 6
Find the tangent to y²=4x at (4,4).
Reveal answer & reasoning
2y=x+4.
The parameter is t=2, so the tangent is ty=x+t².
ORIGINAL PRACTICE · 7
Find the normal at that point.
Reveal answer & reasoning
y=−2x+12.
Use y=−tx+2t+t³ with t=2, giving −2x+4+8.
ORIGINAL PRACTICE · 8
How many real tangents from (2,0) to y²=4x?
Reveal answer & reasoning
None.
The contact equation t²+2=0 has no real roots; equivalently the discriminant is −8.
ORIGINAL PRACTICE · 9
Find the tangent pair from (0,2) to y²=4x.
Reveal answer & reasoning
x=0 and 2y=x+4.
The parameter equation t²−2t=0 gives t=0 and t=2. The first is the vertical vertex tangent.
ORIGINAL PRACTICE · 10
A focal chord of y²=8x has one endpoint parameter 3. Find the other parameter.
Reveal answer & reasoning
−1/3.
The focal chord condition is t₁t₂=−1, independent of the positive scale a.
ORIGINAL PRACTICE · 11
Find the chord joining parameters 1 and −1 on y²=4x.
Reveal answer & reasoning
x=1.
The parameter sum is zero and product −1, so 0=2x−2. It is the vertical latus rectum.
ORIGINAL PRACTICE · 12
Find the contact chord from (−1,2) to y²=4x.
Reveal answer & reasoning
y=x−1.
Use ky=2a(x+h): 2y=2(x−1). The point is exterior because its tangent discriminant is 4+4=8>0.
12 / UNDERSTAND & APPLY
12. Keep the three parameter equations distinct
The point is (at²,2at), the tangent is ty=x+at² and the normal is y=−tx+2at+at³. The tangent-through-point condition is quadratic in t; the normal-through-point condition is cubic. Confusing these structures changes both the line and the number of solutions.
For a chord, use the sum and product of its two parameters. A focal chord has product −1, while the tangent intersection is (at₁t₂,a(t₁+t₂)). These relations explain the directrix locus of perpendicular tangent intersections rather than leaving it as an isolated formula.
The common errors are reading 4a as a, forgetting the opening direction after completing a square, dropping the vertex tangent because its slope is undefined, assuming normals are unique, and treating an axis ray as a finite two-endpoint focal chord.
Check every answer geometrically: a contact point must satisfy the parabola equation, its focus and directrix distances must agree, a tangent must pass through its contact, and a normal must be perpendicular to that tangent. The parameter formulas become dependable when tied back to those defining conditions.
KEEP THE BIG IDEAS
Quick revision
A parabola equates focus distance and directrix distance and has eccentricity 1.
The parameter (at²,2at) covers the entire right-opening parabola.
Implicit tangent and normal formulas include the vertex exceptions.
Focal chords satisfy t₁t₂=−1 and have perpendicular endpoint tangents.
The directrix is the locus of intersections of perpendicular tangent pairs.
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.
Reference inspected for focus–directrix definition, orientations and translated forms. Parameter tangents, normal cubic, focal chords and reflection-vector checks are independently derived with original examples.