Intuition The big pictureA rectangular hyperbola is a hyperbola whose two asymptotes are perpendicular to each other.
The standard 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 = b a=b a = b x 2 − y 2 = a 2 x^2-y^2=a^2 x 2 − y 2 = a 2 rotate the axes by 45 ∘ 45^\circ 4 5 ∘ so that the axes lie along the asymptotes, this equation collapses into the beautifully simple form x y = c 2 xy=c^2 x y = c 2 x y = c 2 xy=c^2 x y = c 2 the same curve as x 2 − y 2 = a 2 x^2-y^2=a^2 x 2 − y 2 = a 2
Definition Rectangular hyperbolaA hyperbola with equal semi-axes (a = b a=b a = b rectangular (or equilateral ). Its eccentricity is e = 2 e=\sqrt{2} e = 2 y = ± x y=\pm x y = ± x
Derivation (from first principles — rotate the axes).
Start with x 2 − y 2 = a 2 x^2-y^2=a^2 x 2 − y 2 = a 2 45 ∘ 45^\circ 4 5 ∘ ( X , Y ) (X,Y) ( X , Y ) ( x , y ) (x,y) ( x , y ) x = X cos 45 ∘ − Y sin 45 ∘ = X − Y 2 , y = X sin 45 ∘ + Y cos 45 ∘ = X + Y 2 x = X\cos45^\circ - Y\sin45^\circ = \frac{X-Y}{\sqrt2},\qquad y = X\sin45^\circ + Y\cos45^\circ = \frac{X+Y}{\sqrt2} x = X cos 4 5 ∘ − Y sin 4 5 ∘ = 2 X − Y , y = X sin 4 5 ∘ + Y cos 4 5 ∘ = 2 X + Y
Why this step? Rotation preserves distances/the shape; we only relabel the same points in turned axes.
Substitute:
x 2 − y 2 = ( X − Y ) 2 2 − ( X + Y ) 2 2 = − 4 X Y 2 = − 2 X Y x^2-y^2 = \frac{(X-Y)^2}{2}-\frac{(X+Y)^2}{2} = \frac{-4XY}{2} = -2XY x 2 − y 2 = 2 ( X − Y ) 2 − 2 ( X + Y ) 2 = 2 − 4 X Y = − 2 X Y − 2 X Y = a 2 -2XY=a^2 − 2 X Y = a 2 X Y = − a 2 2 XY = -\tfrac{a^2}{2} X Y = − 2 a 2 X Y = x ′ y ′ XY=x'y' X Y = x ′ y ′ x y = c 2 , c 2 = a 2 2 \boxed{xy=c^2,\qquad c^2=\frac{a^2}{2}} x y = c 2 , c 2 = 2 a 2
Why this step? It shows x y = c 2 xy=c^2 x y = c 2 not a new curve — it's x 2 − y 2 = a 2 x^2-y^2=a^2 x 2 − y 2 = a 2 x = 0 x=0 x = 0 y = 0 y=0 y = 0
Intuition Why parametrize?On x y = c 2 xy=c^2 x y = c 2 x = c t x=ct x = c t y = c 2 / x = c / t y=c^2/x = c/t y = c 2 / x = c / t t t t
Slope at a point (derive tangent).
Differentiate x y = c 2 xy=c^2 x y = c 2 y + x d y d x = 0 ⇒ d y d x = − y x y + x\frac{dy}{dx}=0 \Rightarrow \dfrac{dy}{dx}=-\dfrac{y}{x} y + x d x d y = 0 ⇒ d x d y = − x y
At ( c t , c / t ) (ct,c/t) ( c t , c / t ) = − c / t c t = − 1 t 2 =-\dfrac{c/t}{ct}=-\dfrac{1}{t^2} = − c t c / t = − t 2 1
Tangent equation (point-slope):
y − c t = − 1 t 2 ( x − c t ) y-\frac{c}{t} = -\frac{1}{t^2}(x-ct) y − t c = − t 2 1 ( x − c t ) t 2 t^2 t 2 t 2 y − c t = − x + c t ⇒ x + t 2 y = 2 c t t^2 y - ct = -x + ct \Rightarrow x + t^2y = 2ct t 2 y − c t = − x + c t ⇒ x + t 2 y = 2 c t x t + t y = 2 c \boxed{\frac{x}{t}+ty = 2c} t x + t y = 2 c
Why this step? Dividing by t t t
Normal equation (slope = + t 2 =+t^2 = + t 2 y − c t = t 2 ( x − c t ) ⇒ x t 3 − y t − c t 4 + c = 0 y-\frac{c}{t}=t^2(x-ct)\ \Rightarrow\ \boxed{xt^3 - yt - ct^4 + c = 0} y − t c = t 2 ( x − c t ) ⇒ x t 3 − y t − c t 4 + c = 0
Worked example 1) Tangent att = 2 t=2 t = 2 x y = 9 xy=9 x y = 9 c 2 = 9 ⇒ c = 3 c^2=9\Rightarrow c=3 c 2 = 9 ⇒ c = 3 ( c t , c / t ) = ( 6 , 1.5 ) (ct,c/t)=(6,\ 1.5) ( c t , c / t ) = ( 6 , 1.5 ) x t + t y = 2 c ⇒ x 2 + 2 y = 6 ⇒ x + 4 y = 12. \frac{x}{t}+ty=2c \Rightarrow \frac{x}{2}+2y=6 \Rightarrow x+4y=12. t x + t y = 2 c ⇒ 2 x + 2 y = 6 ⇒ x + 4 y = 12. Why this step? Plug c = 3 , t = 2 c=3,t=2 c = 3 , t = 2 6 + 4 ( 1.5 ) = 12 6+4(1.5)=12 6 + 4 ( 1.5 ) = 12
Worked example 2) Point of intersection of tangents att 1 , t 2 t_1,t_2 t 1 , t 2 x t 1 + t 1 y = 2 c \frac{x}{t_1}+t_1y=2c t 1 x + t 1 y = 2 c x t 2 + t 2 y = 2 c \frac{x}{t_2}+t_2y=2c t 2 x + t 2 y = 2 c x ( 1 t 1 − 1 t 2 ) + y ( t 1 − t 2 ) = 0 x\left(\frac1{t_1}-\frac1{t_2}\right)+y(t_1-t_2)=0 x ( t 1 1 − t 2 1 ) + y ( t 1 − t 2 ) = 0 1 t 1 − 1 t 2 = t 2 − t 1 t 1 t 2 \frac1{t_1}-\frac1{t_2}=\frac{t_2-t_1}{t_1t_2} t 1 1 − t 2 1 = t 1 t 2 t 2 − t 1 − x t 1 t 2 + y = 0 ⇒ x = t 1 t 2 y -\frac{x}{t_1t_2}+y=0\Rightarrow x=t_1t_2\,y − t 1 t 2 x + y = 0 ⇒ x = t 1 t 2 y t 1 t 2 y t 1 + t 1 y = 2 c ⇒ y ( t 2 + t 1 ) = 2 c ⇒ y = 2 c t 1 + t 2 \frac{t_1t_2y}{t_1}+t_1y=2c\Rightarrow y(t_2+t_1)=2c\Rightarrow y=\frac{2c}{t_1+t_2} t 1 t 1 t 2 y + t 1 y = 2 c ⇒ y ( t 2 + t 1 ) = 2 c ⇒ y = t 1 + t 2 2 c ( 2 c t 1 t 2 t 1 + t 2 , 2 c t 1 + t 2 ) \boxed{\left(\frac{2ct_1t_2}{t_1+t_2},\ \frac{2c}{t_1+t_2}\right)} ( t 1 + t 2 2 c t 1 t 2 , t 1 + t 2 2 c ) Why this step? Elegant symmetric answer used constantly in JEE chord problems.
Worked example 3) Chord joiningt 1 , t 2 t_1,t_2 t 1 , t 2 ( c t 1 , c / t 1 ) , ( c t 2 , c / t 2 ) (ct_1,c/t_1),(ct_2,c/t_2) ( c t 1 , c / t 1 ) , ( c t 2 , c / t 2 ) = c / t 1 − c / t 2 c t 1 − c t 2 = ( t 2 − t 1 ) / ( t 1 t 2 ) t 1 − t 2 = − 1 t 1 t 2 =\dfrac{c/t_1-c/t_2}{ct_1-ct_2}=\dfrac{(t_2-t_1)/(t_1t_2)}{t_1-t_2}=-\dfrac{1}{t_1t_2} = c t 1 − c t 2 c / t 1 − c / t 2 = t 1 − t 2 ( t 2 − t 1 ) / ( t 1 t 2 ) = − t 1 t 2 1 y − c t 1 = − 1 t 1 t 2 ( x − c t 1 ) y-\frac{c}{t_1}=-\frac1{t_1t_2}(x-ct_1) y − t 1 c = − t 1 t 2 1 ( x − c t 1 ) x + t 1 t 2 y = c ( t 1 + t 2 ) \boxed{x + t_1t_2\,y = c(t_1+t_2)} x + t 1 t 2 y = c ( t 1 + t 2 ) Why this step? Setting t 1 = t 2 = t t_1=t_2=t t 1 = t 2 = t x + t 2 y = 2 c t x+t^2y=2ct x + t 2 y = 2 c t
Worked example 4) Rectangle property (area constant)From any point ( c t , c / t ) (ct,c/t) ( c t , c / t ) ∣ c t ∣ |ct| ∣ c t ∣ ∣ c / t ∣ |c/t| ∣ c / t ∣ = c t ⋅ c t = c 2 =ct\cdot\frac{c}{t}=c^2 = c t ⋅ t c = c 2 constant, independent of t t t
Why this matters: This is the geometric reason the curve is x y = c 2 xy=c^2 x y = c 2
Common mistake Steel-man the common errors(a) "x y = c 2 xy=c^2 x y = c 2
Why it feels right: every standard hyperbola we saw had a horizontal or vertical axis. Fix: Here the axis is the line y = x y=x y = x asymptotes , not the axes of the curve.
(b) Slope of tangent taken as + 1 / t 2 +1/t^2 + 1/ t 2
Why it feels right: people forget the minus in d y / d x = − y / x dy/dx=-y/x d y / d x = − y / x Fix: y y y x x x negative = − 1 / t 2 =-1/t^2 = − 1/ t 2 normal is + t 2 +t^2 + t 2
(c) Using x a 2 \frac{x}{a^2} a 2 x
Why it feels right: muscle memory from ellipse/standard hyperbola. Fix: x y = c 2 xy=c^2 x y = c 2 x t + t y = 2 c \frac{x}{t}+ty=2c t x + t y = 2 c
Recall Feynman: explain to a 12-year-oldImagine two perfectly straight roads that cross at right angles (the x and y axes). Now draw a curve that hugs both roads but never touches them, in two opposite corners. Wherever you stand on the curve, the box you make back to the two roads always has the same area — that fixed area is c 2 c^2 c 2 is the curve x y = c 2 xy=c^2 x y = c 2
Mnemonic Remember the tangent"Divide the point-index, add — equals two-c."
x t + t y = 2 c \dfrac{x}{t}+ty=2c t x + t y = 2 c 1 t \frac1t t 1 x x x t t t y y y 2 c 2c 2 c Flip the signs of the reciprocals into cubes: x t 3 − y t = c ( t 4 − 1 ) xt^3-yt=c(t^4-1) x t 3 − y t = c ( t 4 − 1 )
What makes a hyperbola "rectangular"? Its asymptotes are perpendicular (equivalently
a = b a=b a = b ,
e = 2 e=\sqrt2 e = 2 ).
Eccentricity of a rectangular hyperbola? Standard form of rectangular hyperbola referred to its asymptotes? Parametric point on x y = c 2 xy=c^2 x y = c 2 ( c t , c / t ) (ct,\ c/t) ( c t , c / t ) .
d y / d x dy/dx d y / d x x y = c 2 xy=c^2 x y = c 2 − y / x -y/x − y / x ; at
( c t , c / t ) (ct,c/t) ( c t , c / t ) it is
− 1 / t 2 -1/t^2 − 1/ t 2 .
Tangent to x y = c 2 xy=c^2 x y = c 2 t t t x t + t y = 2 c \dfrac{x}{t}+ty=2c t x + t y = 2 c .
Normal to x y = c 2 xy=c^2 x y = c 2 t t t x t 3 − y t = c ( t 4 − 1 ) xt^3 - yt = c(t^4-1) x t 3 − y t = c ( t 4 − 1 ) .
Chord joining t 1 , t 2 t_1,t_2 t 1 , t 2 x y = c 2 xy=c^2 x y = c 2 x + t 1 t 2 y = c ( t 1 + t 2 ) x+t_1t_2\,y=c(t_1+t_2) x + t 1 t 2 y = c ( t 1 + t 2 ) .
Intersection of tangents at t 1 , t 2 t_1,t_2 t 1 , t 2 ( 2 c t 1 t 2 t 1 + t 2 , 2 c t 1 + t 2 ) \left(\dfrac{2ct_1t_2}{t_1+t_2},\ \dfrac{2c}{t_1+t_2}\right) ( t 1 + t 2 2 c t 1 t 2 , t 1 + t 2 2 c ) .
Constant-area property of x y = c 2 xy=c^2 x y = c 2 Rectangle from any point to the axes has area
c 2 c^2 c 2 .
Vertices of x y = c 2 xy=c^2 x y = c 2 ( c , c ) (c,c) ( c , c ) and
( − c , − c ) (-c,-c) ( − c , − c ) on line
y = x y=x y = x .
x 2 − y 2 = a 2 x^2-y^2=a^2 x 2 − y 2 = a 2 x y = c 2 xy=c^2 x y = c 2 Rotate axes
45 ∘ 45^\circ 4 5 ∘ ; then
c 2 = a 2 / 2 c^2=a^2/2 c 2 = a 2 /2 .
Hyperbola standard form — x y = c 2 xy=c^2 x y = c 2 a = b a=b a = b 45 ∘ 45^\circ 4 5 ∘ Rotation of axes — the tool that converts x 2 − y 2 = a 2 → x y = c 2 x^2-y^2=a^2 \to xy=c^2 x 2 − y 2 = a 2 → x y = c 2 Asymptotes of a hyperbola — here they become the coordinate axes.Tangent and Normal to conics — parametric method shown above.Eccentricity — why e = 2 e=\sqrt2 e = 2 Reciprocal function y=1/x — the graph y = c 2 / x y=c^2/x y = c 2 / x
Standard hyperbola x2/a2 - y2/b2 = 1
Rectangular hyperbola a=b
Asymptotes x=0 and y=0 perpendicular
Slope dy/dx = -y/x = -1/t2
Normal xt3 - yt - ct4 + c = 0
Intuition Hinglish mein samjhoDekho, rectangular hyperbola matlab aisa hyperbola jiske dono asymptotes ek dusre ke perpendicular hote hain. Standard 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 = b a=b a = b x 2 − y 2 = a 2 x^2-y^2=a^2 x 2 − y 2 = a 2 e = 2 e=\sqrt2 e = 2
Ab magic ye hai: agar hum axes ko 45 ∘ 45^\circ 4 5 ∘ x y = c 2 xy=c^2 x y = c 2 c 2 = a 2 / 2 c^2=a^2/2 c 2 = a 2 /2 x y = c 2 xy=c^2 x y = c 2 y = x y=x y = x
Sabse kaam ki cheez hai parametric point ( c t , c / t ) (ct, c/t) ( c t , c / t ) t t t x t + t y = 2 c \frac{x}{t}+ty=2c t x + t y = 2 c − 1 / t 2 -1/t^2 − 1/ t 2 x t 3 − y t = c ( t 4 − 1 ) xt^3-yt=c(t^4-1) x t 3 − y t = c ( t 4 − 1 ) t 1 , t 2 t_1,t_2 t 1 , t 2 x + t 1 t 2 y = c ( t 1 + t 2 ) x+t_1t_2y=c(t_1+t_2) x + t 1 t 2 y = c ( t 1 + t 2 ) t 1 = t 2 t_1=t_2 t 1 = t 2
Geometric feel: curve ke kisi bhi point se dono axes tak perpendicular giraao, jo rectangle banega uska area hamesha c 2 c^2 c 2 2 c 2c 2 c