Aims tutorial 1 st year –Mathematics IB (Important questions) 1. When the origin is shifted to the point (2, 3),the transformed equation of the curve isX2+3 XY−2Y 2+17 X−7Y−11=0. Find the original equation of the curve.sol : Transformed equation of the curve is

X2+3 XY−2Y 2+17 X−7Y−11=0Here X=x-h=x-2 and Y=y-k=y-3Original eq’’n is(x-2)2 + 3(x-2)(y-3)-2(y-3)2+17(x-2) – 7(y-3)-11=0⇒x2-4x+4+3xy-9x-6y+18-2y2+12y-18+17x-34-7y+21-11=0⇒x2+3xy-2y2+4x-y-20=0.2. When the origin is shifted to the point (-1, 2), by the translation of axes. Find the transformed equation ofx2+ y2+2x−4 y+1=0.Sol: origin equation of the curve isx2+ y2+2x−4 y+1=0.herex=X+h=X-1 and y=Y+k=Y+2(X−1)2+(Y +2)2+2(X−1)−4(Y +2)+1=0.⇒X2+1-2X+Y2+4+4Y+2X-2-4Y-8+1=0.∴ X2+Y 2−4=0 .3. When the axes are rotated through an angle 300,find the transformed equation of x2+2√3 xy− y2=2a2.sol :x=Xcosθ-Ysinθ and y=Xsinθ + Ycosθ

givenθ=30 Sin30=1

2and cos30=√3

2

⇒ x=X √32

−Y2=√3 X−Y

2

⇒y= X2

+ y √32

= X+√3Y2Transformed equation is

⇒ [√3 X−Y2

]2+2√ 3[√3 X−Y

2]. [X+√3Y

2

]-[ X+√3Y2

]2=2a2

⇒ 3 X2−2√3 XY +Y 2

4+ 2√3 [√3 X

2−XY+3 XY−√ 3Y 2]4

−[ X2+3Y 2+2√3 XY

4]=2a2

3 X2−2√3 XY+Y 2+6 X2+4 √3 XY−6Y 2−X2−3Y 2−2√3 XY4

=2a2

⇒8X2-8Y2=8a2∴X2 – Y2 =a2

4. When the axes are rotated through an angle 450,find the transformed equation of 3 x2+10 xy+3 y2=9.H /W5. When the axes are rotated through an angle 450, the transformed equation of the curve is

17 x2−16 xy+17 y2=225.find the original equation of the curve.Sol: X=xcosθ+ ysinθ Y=-xsinθ+ ycosθGiven θ=45 Sin45= 1

√2 and cos30= 1

√2

x=x 1√ 2

+ y 1√2

= x+ y√2

⇒y=−x√2

+ y

√2=−x+ y

√2the original equation of the curve is ⇒17X2-16XY+17Y2=0⇒17[ x+ y√2

]2

−16 [ x+ y√2 ] [−x+ y√ 2 ]+17[−x+ y√2]2

=225

⇒17[ x2+ y2+2 xy ]

2−16[ y

2−x2]2

+ 17 [x2+ y2−2 xy ]2

=225

⇒17 x2+17 y2+34 xy−16 y2+16 x2+17 x2+17 y2−34 xy2=225

⇒50 x2+18 y2=450∴25 x2+9 y2=225 Or x2

9+ y

2

25=1 is the original

eq’’n.6. When the axes are rotated through an angle α , find the transformed equation of xcos α+ ysinα=p .Sol: the given eq’’n is xcos α+ ysinα=p .∴ the axes are rotated through an angle α⇒x=Xcosα−Ysinα∧ y=Xsinα+Ycosαthe transformed equation is (Xcos α−Ysinα)cosα+(Xsinα+Ycosα)sinα=p .⇒Xcos2α -Ycosα sinα+Xsin2α+Ycosαsinα=p⇒ Xcos2α + Xsin2α=p⇒X (cos2α + sin2α )=p∴ the transformed eq’’n is X=p.

Aims tutorial 1 st year –Mathematics IB (Important questions) 7. Show that the axes are rotated through an

angle of 12tan−1[ 2ha−b ] so as to remove the

xy terms from the equation ax2+2hxy+b y2=0 ,if a≠b and through the angleπ

4, if a=b.

Sol: If the axes are rotated through an angle ' α ,then⇒ x=Xcosα−Ysinα∧ y=Xsinα+YcosαTherefore the given eq’’n transformed as ⇒a(Xcos α−Ysinα)2+ 2h [Xcosα−Ysinα][Xsinα+Ycosα]+b(Xsinα+Ycosα )2=0To remove XY terms from the eq’’n, the coeff of XY term must be zero.-2XYcosα . sinα +2hXYcos2α -2hXYsin2α+2bXycosα .sinα=0⇒i.e, (b-a) sinα .cosα + h(cos2α−sin 2α)=0

⇒hcos2α=(a−b)2

sin 2α

⇒tan2α= 2ha−b

⇒α=1

2tan−1

2ha−b If a=b ⇒hcos2α=0⇒2α=90∴α=45