1.

If =Xcostheta-Ysintheta,q=Xsintheta+Ycosthetaandp^(2)+4pq+q^(2)=AX^(2)+BY^(2),0le0le(pi)/(2).What is the value of A ?

Answer»

4
3
2
1

Solution :`p=xcostheta-ysintheta`
`Q=xsintheta+ycostheta`
Given, `p^(2)+4pq+q^(2)=Ax^(2)+By^(2)`
Letus take`theta=(pi)/(4)`.
`p=X"COS"(pi)/(4)-y" cos"(pi)/(4)=(x+y)/(sqrt(2))`
`q=x" SIN"(pi)/(4)+y" cos"(pi)/(4)=(x+y)/(sqrt(2))`
`pq=(x^(2)-y^(2))/(2)rArr2pq=x^(2)-y^(2)rArr4pq=2x^(2)-2y^(2)` . . . (1)
Now, `p^(2)+q^(2)=x^(2)cos^(2)theta+y^(2)sin^(2)theta-2xycosthetasintheta+x^(2)sin^(2)theta+y^(2)cos^(2)theta+2xysinthetacostheta=x^(2)+y^(2)`. . . (2)
From(1) , (2) , `p^(2)+q^(2)+4pq=x^(2)+y^(2)+2x^(2)-2y^(2)=3x^(2)-y^(2)`
Comparingthis with the given from , we get
`theta=(pi)/(4),A=3,B=-1`


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