Let `P -=(x,y)`. According to the question,
`x=acostheta`(1)
`y=bsintheta` (2)
Sqaureing and adding (1) and (2), we get
`(x^2)/(a^2)+(y^2)/(b^2)=cos^2theta+sin^2theta`
or `(x^2)/(a^2)+(y^2)/(b^2)=1`
The direction cosines of `vec(OP)` are `-(1)/(3), (2)/(3) and -(2)/(3)`.
Hence, `vec(OP)=|vec(OP)|(lhati+mhatj+nhatk)`
`" "=3(-(1)/(3)hati+(2)/(3)hatj-(2)/(3)hatk)`
`" "=-1hati+2hatj-2hatk`
So, the coordinations of P are -1, 2 and -2.
By defing A & B are equal if they have the same order and all the corresponding elements are equal.
Thus we have `sin theta=(1)/(sqrt2),c os theta=-(1)/(sqrt2)& tan theta=-1`
`Rightarrow...
Correct Answer - C
`AB=[(cos^(2) theta,),(cos theta sin theta,)][(cos^(2) phi,cos phi sin phi),(cos phi sin phi,sin^(2) phi)]`
`=[(cos^(2) theta cos^(2) phi+cos theta cos phi sin theta sin phi ,cos^(2)theta cos phi...
Correct Answer - A::B::C
We have, `|A(theta)|=1`
Hence, A is invertiable.
`A(pi+theta)=A(pi + theta)=[(sin(pi+theta),i cos (pi+theta)),(i cos (pi+theta),sin (pi+theta))]`
`=[(-sin theta,-i co theta),(-i cos theta,-sin theta)]=-A(theta)`
adj `(A(theta))=[(sin theta,-i cos theta),(-i...
Consider any point P`(a sec theta, a tan theta)" on "x^(2)-y^(2)=a^(2)` This point will be nearest to y = 2x if the tangent at this point is parallel to y...
Chord of contact of parabola `y^(2)=4ax` w.r.t. point `P(x_(1),y_(1))` is `yy_(1)=2a(x+x_(1))`.
Solving this line with parabola `x^(2)=4by`, we get
`x^(2)=4bxx(2a)/(y_(1))(x+x_(1))`
`ory_(1)x^(2)=8bx-8abx_(1)=0`
Since the line touches the parabola, this equation will...