A conic is a set of points satisfying one Cartesian equation (like x 2 a 2 + y 2 b 2 = 1 \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 a 2 x 2 + b 2 y 2 = 1 walk along the curve — plug in one number t t t θ \theta θ ( x , y ) (x,y) ( x , y ) guaranteed to lie on the curve. That single-number description is a parametric form .
WHY it helps: one variable instead of a constraint between two. Tangents, chords, loci, and integration all become algebra in t t t instead of implicit differentiation.
Definition ParametrizationA parametric form of a curve is a pair of functions x = f ( t ) , y = g ( t ) x=f(t),\ y=g(t) x = f ( t ) , y = g ( t ) t t t ( f ( t ) , g ( t ) ) (f(t),g(t)) ( f ( t ) , g ( t )) identically (for all t t t
The magic tool is a trig / hyperbolic identity whose shape matches the conic's equation.
Identity
Matches conic
cos 2 θ + sin 2 θ = 1 \cos^2\theta+\sin^2\theta=1 cos 2 θ + sin 2 θ = 1 ellipse / circle
sec 2 θ − tan 2 θ = 1 \sec^2\theta-\tan^2\theta=1 sec 2 θ − tan 2 θ = 1 hyperbola
y 2 = x ⋅ ( … ) y^2=x\cdot(\dots) y 2 = x ⋅ ( … ) parabola
We need x , y x,y x , y x 2 + y 2 x^2+y^2 x 2 + y 2 r 2 r^2 r 2 cos 2 θ + sin 2 θ = 1 \cos^2\theta+\sin^2\theta=1 cos 2 θ + sin 2 θ = 1 begging to be used. Multiply it by r 2 r^2 r 2 r 2 cos 2 θ + r 2 sin 2 θ = r 2 . r^2\cos^2\theta+r^2\sin^2\theta=r^2. r 2 cos 2 θ + r 2 sin 2 θ = r 2 . x 2 + y 2 = r 2 x^2+y^2=r^2 x 2 + y 2 = r 2
The Cartesian equation is already "( ⋅ ) 2 + ( ⋅ ) 2 = 1 (\cdot)^2+(\cdot)^2=1 ( ⋅ ) 2 + ( ⋅ ) 2 = 1 x a = cos θ , y b = sin θ . \frac{x}{a}=\cos\theta,\quad \frac{y}{b}=\sin\theta. a x = cos θ , b y = sin θ . Why this step? Then x 2 a 2 + y 2 b 2 = cos 2 θ + sin 2 θ = 1 \frac{x^2}{a^2}+\frac{y^2}{b^2}=\cos^2\theta+\sin^2\theta=1 a 2 x 2 + b 2 y 2 = cos 2 θ + sin 2 θ = 1
Now we have a minus . We need an identity of the form ( ⋅ ) 2 − ( ⋅ ) 2 = 1 (\cdot)^2-(\cdot)^2=1 ( ⋅ ) 2 − ( ⋅ ) 2 = 1 sec 2 θ − tan 2 θ = 1 \sec^2\theta-\tan^2\theta=1 sec 2 θ − tan 2 θ = 1 x a = sec θ , y b = tan θ . \frac{x}{a}=\sec\theta,\quad \frac{y}{b}=\tan\theta. a x = sec θ , b y = tan θ .
No sum/difference identity needed — the equation directly links x x x y 2 y^2 y 2 y y y t t t y = 2 a t y=2at y = 2 a t y 2 = 4 a 2 t 2 = 4 a x ⟹ x = a t 2 . y^2=4a^2t^2=4ax \implies x=at^2. y 2 = 4 a 2 t 2 = 4 a x ⟹ x = a t 2 . Why y = 2 a t y=2at y = 2 a t y = t y=t y = t We chose the neat constant 2 a 2a 2 a x x x a t 2 at^2 a t 2
Worked example E1 — Find a point & check it lies on the ellipseEllipse x 2 25 + y 2 9 = 1 \frac{x^2}{25}+\frac{y^2}{9}=1 25 x 2 + 9 y 2 = 1 θ = π 3 \theta=\frac{\pi}{3} θ = 3 π x = 5 cos π 3 = 5 ⋅ 1 2 = 5 2 x=5\cos\frac{\pi}{3}=5\cdot\frac12=\frac52 x = 5 cos 3 π = 5 ⋅ 2 1 = 2 5 y = 3 sin π 3 = 3 ⋅ 3 2 = 3 3 2 y=3\sin\frac{\pi}{3}=3\cdot\frac{\sqrt3}{2}=\frac{3\sqrt3}{2} y = 3 sin 3 π = 3 ⋅ 2 3 = 2 3 3 Check: ( 5 / 2 ) 2 25 + ( 3 3 / 2 ) 2 9 = 25 / 4 25 + 27 / 4 9 = 1 4 + 3 4 = 1. \frac{(5/2)^2}{25}+\frac{(3\sqrt3/2)^2}{9}=\frac{25/4}{25}+\frac{27/4}{9}=\frac14+\frac34=1. 25 ( 5/2 ) 2 + 9 ( 3 3 /2 ) 2 = 25 25/4 + 9 27/4 = 4 1 + 4 3 = 1. Why this step? Plugging back into the Cartesian equation is the definition of "lies on the curve".
Worked example E2 — Tangent to a parabola using the parameterFind the tangent to y 2 = 4 a x y^2=4ax y 2 = 4 a x t t t ( a t 2 , 2 a t ) (at^2,2at) ( a t 2 , 2 a t ) d x d t = 2 a t , d y d t = 2 a \frac{dx}{dt}=2at,\ \frac{dy}{dt}=2a d t d x = 2 a t , d t d y = 2 a = d y / d t d x / d t = 2 a 2 a t = 1 t =\frac{dy/dt}{dx/dt}=\frac{2a}{2at}=\frac1t = d x / d t d y / d t = 2 a t 2 a = t 1 y − 2 a t = 1 t ( x − a t 2 ) ⇒ t y − 2 a t 2 = x − a t 2 ⇒ t y = x + a t 2 y-2at=\frac1t(x-at^2)\Rightarrow ty-2at^2=x-at^2\Rightarrow ty=x+at^2 y − 2 a t = t 1 ( x − a t 2 ) ⇒ t y − 2 a t 2 = x − a t 2 ⇒ t y = x + a t 2 t y = x + a t 2 \boxed{ty=x+at^2} t y = x + a t 2 Why this step? Parametric differentiation avoids messy implicit work — slope pops out as 1 / t 1/t 1/ t
Worked example E3 — Chord joining two points on a hyperbolaOn x 2 a 2 − y 2 b 2 = 1 \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 a 2 x 2 − b 2 y 2 = 1 P ( a sec α , b tan α ) P(a\sec\alpha,b\tan\alpha) P ( a sec α , b tan α ) Q ( a sec β , b tan β ) Q(a\sec\beta,b\tan\beta) Q ( a sec β , b tan β ) = b tan α − b tan β a sec α − a sec β =\dfrac{b\tan\alpha-b\tan\beta}{a\sec\alpha-a\sec\beta} = a sec α − a sec β b tan α − b tan β tan = sin cos \tan=\frac{\sin}{\cos} tan = c o s s i n β → α \beta\to\alpha β → α Why this step? Two-parameter chords are the gateway to tangent/normal formulae; the whole machinery lives in the parameter.
Worked example E4 — Convert back (eliminate the parameter)Given x = 3 cos θ , y = 2 sin θ x=3\cos\theta,\ y=2\sin\theta x = 3 cos θ , y = 2 sin θ cos θ = x 3 , sin θ = y 2 \cos\theta=\frac x3,\ \sin\theta=\frac y2 cos θ = 3 x , sin θ = 2 y cos 2 + sin 2 = 1 \cos^2+\sin^2=1 cos 2 + sin 2 = 1 x 2 9 + y 2 4 = 1 \frac{x^2}{9}+\frac{y^2}{4}=1 9 x 2 + 4 y 2 = 1 Why this step? Eliminating θ \theta θ proves the parametric form is correct — reverses the derivation.
Common mistake "For an ellipseθ \theta θ Why it feels right: for a circle θ \theta θ The fix: the ellipse point is the circle point pushed vertically by factor b / a b/a b / a θ \theta θ eccentric angle on the auxiliary circle, not arctan ( y / x ) \arctan(y/x) arctan ( y / x )
x = a cos θ , y = b sin θ x=a\cos\theta,\ y=b\sin\theta x = a cos θ , y = b sin θ hyperbola .
Why it feels right: it's the same look as the ellipse. The fix: the hyperbola has a minus sign — you need an identity with a minus, i.e. sec 2 − tan 2 = 1 \sec^2-\tan^2=1 sec 2 − tan 2 = 1
Common mistake Writing parabola as( t 2 , 2 t ) (t^2,2t) ( t 2 , 2 t ) a a a Why it feels right: you memorised the shape "( t 2 , 2 t ) (t^2,2t) ( t 2 , 2 t ) The fix: the a a a ( a t 2 , 2 a t ) (at^2,2at) ( a t 2 , 2 a t ) y 2 = 4 a 2 t 2 = 4 a ⋅ a t 2 = 4 a x y^2=4a^2t^2=4a\cdot at^2=4ax y 2 = 4 a 2 t 2 = 4 a ⋅ a t 2 = 4 a x
Common mistake Forgetting the hyperbola trig form misses a branch.Why it feels right: you got infinitely many points. The fix: sec θ \sec\theta sec θ θ \theta θ π 2 \frac\pi2 2 π cosh t ≥ 1 \cosh t\ge1 cosh t ≥ 1 right branch — use ± a cosh t \pm a\cosh t ± a cosh t
Recall Feynman: explain to a 12-year-oldImagine a curve is a racetrack. Instead of describing every point by "how far right and how far up," we give a stopwatch time t t t t = 3 t=3 t = 3 angle you've turned. For a parabola it's just a plain number. One number tells the whole position — that's a parametric form.
Mnemonic Which identity for which conic?"Circles & ellipses are SweetCorn (sin / cos \sin/\cos sin / cos a t 2 , 2 a t at^2,2at a t 2 , 2 a t
Sum of squares = = = = = =
Conic
Cartesian
Parametric
Identity used
Range
Circle
x 2 + y 2 = r 2 x^2+y^2=r^2 x 2 + y 2 = r 2 ( r cos θ , r sin θ ) (r\cos\theta, r\sin\theta) ( r cos θ , r sin θ ) c 2 + s 2 = 1 c^2+s^2=1 c 2 + s 2 = 1 [ 0 , 2 π ) [0,2\pi) [ 0 , 2 π )
Ellipse
x 2 a 2 + y 2 b 2 = 1 \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 a 2 x 2 + b 2 y 2 = 1 ( a cos θ , b sin θ ) (a\cos\theta, b\sin\theta) ( a cos θ , b sin θ ) c 2 + s 2 = 1 c^2+s^2=1 c 2 + s 2 = 1 [ 0 , 2 π ) [0,2\pi) [ 0 , 2 π )
Parabola
y 2 = 4 a x y^2=4ax y 2 = 4 a x ( a t 2 , 2 a t ) (at^2, 2at) ( a t 2 , 2 a t ) direct
R \mathbb R R
Hyperbola
x 2 a 2 − y 2 b 2 = 1 \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 a 2 x 2 − b 2 y 2 = 1 ( a sec θ , b tan θ ) (a\sec\theta, b\tan\theta) ( a sec θ , b tan θ ) sec 2 − tan 2 = 1 \sec^2-\tan^2=1 sec 2 − tan 2 = 1 θ ≠ π 2 \theta\ne\frac\pi2 θ = 2 π
Hyperbola (alt)
same
( a cosh t , b sinh t ) (a\cosh t, b\sinh t) ( a cosh t , b sinh t ) cosh 2 − sinh 2 = 1 \cosh^2-\sinh^2=1 cosh 2 − sinh 2 = 1 right branch
Conic Sections - Standard equations Eccentric angle and auxiliary circle Tangents and normals to conics Chord of contact and pole-polar Trigonometric identities Hyperbolic functions Parametric differentiation
#flashcards/maths
Parametric form of the circle x 2 + y 2 = r 2 x^2+y^2=r^2 x 2 + y 2 = r 2 x = r cos θ , y = r sin θ x=r\cos\theta,\ y=r\sin\theta x = r cos θ , y = r sin θ Which identity powers the ellipse parametrization? cos 2 θ + sin 2 θ = 1 \cos^2\theta+\sin^2\theta=1 cos 2 θ + sin 2 θ = 1 Parametric form of ellipse x 2 a 2 + y 2 b 2 = 1 \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 a 2 x 2 + b 2 y 2 = 1 x = a cos θ , y = b sin θ x=a\cos\theta,\ y=b\sin\theta x = a cos θ , y = b sin θ Why can't ( a cos θ , b sin θ ) (a\cos\theta,b\sin\theta) ( a cos θ , b sin θ ) Hyperbola has a minus sign; need
sec 2 − tan 2 = 1 \sec^2-\tan^2=1 sec 2 − tan 2 = 1 , not
cos 2 + sin 2 \cos^2+\sin^2 cos 2 + sin 2 Parametric form of hyperbola x 2 a 2 − y 2 b 2 = 1 \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 a 2 x 2 − b 2 y 2 = 1 x = a sec θ , y = b tan θ x=a\sec\theta,\ y=b\tan\theta x = a sec θ , y = b tan θ Alternative hyperbolic parametrization of the hyperbola x = a cosh t , y = b sinh t x=a\cosh t,\ y=b\sinh t x = a cosh t , y = b sinh t (right branch)
Parametric form of parabola y 2 = 4 a x y^2=4ax y 2 = 4 a x x = a t 2 , y = 2 a t x=at^2,\ y=2at x = a t 2 , y = 2 a t What is θ \theta θ The eccentric angle
Tangent to y 2 = 4 a x y^2=4ax y 2 = 4 a x t t t t y = x + a t 2 ty=x+at^2 t y = x + a t 2 Slope of parabola at point t t t ( a t 2 , 2 a t ) (at^2,2at) ( a t 2 , 2 a t ) How do you verify a parametric form? Eliminate the parameter and recover the Cartesian equation identically
Parametric form x=f t y=g t
Cartesian equation identically
Trig / hyperbolic identity
Hyperbola x=a sec, y=b tan
Intuition Hinglish mein samjhoDekho, har conic ek Cartesian equation se define hoti hai — jaise ellipse ke liye x 2 a 2 + y 2 b 2 = 1 \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 a 2 x 2 + b 2 y 2 = 1 upar chalna hota hai, matlab ek hi number t t t θ \theta θ ( x , y ) (x,y) ( x , y ) parametric form — do variables ki jagah ek variable se poori curve.
Trick simple hai: aisi ek trig identity dhoondo jiska shape conic ki equation se match kare . Sum of squares ( cos 2 + sin 2 = 1 ) (\cos^2+\sin^2=1) ( cos 2 + sin 2 = 1 ) x = a cos θ , y = b sin θ x=a\cos\theta,\ y=b\sin\theta x = a cos θ , y = b sin θ sec 2 − tan 2 = 1 \sec^2-\tan^2=1 sec 2 − tan 2 = 1 x = a sec θ , y = b tan θ x=a\sec\theta,\ y=b\tan\theta x = a sec θ , y = b tan θ y = 2 a t y=2at y = 2 a t x = a t 2 x=at^2 x = a t 2
Ek important yaad rakhne wali baat: ellipse mein jo θ \theta θ seedha angle nahi hai — usko eccentric angle kehte hain, jo auxiliary circle (radius a a a cos , sin \cos,\sin cos , sin sec , tan \sec,\tan sec , tan
Yeh forms exam mein gold hain: tangent, normal, chord, aur locus ke saare tough sawaal parameter t t t t t t t y = x + a t 2 ty=x+at^2 t y = x + a t 2
