# Determine the type of conic, remove the xy term by rotation of axes, reduce the new equation to a standard form and trace the curve on the new axes. 1.) 3x^2 -4xy + 3y^2 =10 2.) 7x^2 - 8xy + 13y^2...

Determine the type of conic, remove the xy term by rotation of axes, reduce the new equation to a standard form and trace the curve on the new axes.

1.) 3x^2 -4xy + 3y^2 =10

2.) 7x^2 - 8xy + 13y^2 -45 = 0

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### 1 Answer

Qest1. 3x^2-4xy+3y^2=10 (i)

A=3=c , B=-4

rotate the axes so the conic is parallel to one of the axes.

`tan(2theta)=B/(A-C)=-4/0`

```2theta=90 ,theta=45^o`

`x=x'cos(theta)-y'sin(theta)`

`y=x'sin(theta)+y'cos(theta)`

`x=(1/sqrt(2))(x'-y')`

`y=(1/sqrt(2))(x'+y')`

``Substituting the value of x and y ,and simplify the equation ,we have

`3((x'-y')/sqrt(2))^2-4(x'-y')/sqrt(2)(x'+y')/sqrt(2)+3((x'+y')/sqrt(2))^2=10`

`7y'^2-x'^2=10`

`(y'^2)/(10/7)-(x'^2)/(10/1)=1`

which is hyperbola and its transverse axis is along axis of y'.

Qest 2.

7x^2 - 8xy + 13y^2 -45 = 0

7x^2-8xy+13y^2=45

A=7 ,C=13, B=-8

Tan(2theta)=-8/(7-13) , x=x'cos(theta)-y'sin(theta),y=x'sin(theta)+y'cos(theta)

theta=26.5650

substituting values in given equation

(7-4sin(2 theta)+6sin^2(theta))x^2+(7+4sin(2 theta)+6cos^2(theta))y^2=45

which is ellipse.