The distance of a point -6cos- -2sin- on the ellipse x26 - y22-1 from the centre is 2- if-

The correct option is D θ = #160;π4 P(√( 6 )cosθ #160; ,√( 2 )sinθ #160; ) Centre(0,0) PC=√( 6cos ^ 2 θ #160; +2sin ^ 2 θ #160; ...

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Standard XII

Mathematics

Evaluation of a Determinant

The distance ...

Question

The distance of a point $(\sqrt{6} cos θ, \sqrt{2} sin θ)$ on the ellipse $\frac{x^{2}}{6} + \frac{y^{2}}{2} = 1$ from the centre is $2$ , if:

θ = \frac{π}{2}

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θ = \frac{3 π}{2}

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θ = \frac{5 π}{2}

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θ = \frac{π}{4}

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Solution

The correct option is D $θ = \frac{π}{4}$
$P (\sqrt{6} cos θ, \sqrt{2} sin θ)$
$C e n t r e (0, 0)$
$P C = \sqrt{6 {cos}^{2} θ + 2 {sin}^{2} θ}$
$2 = \sqrt{4 {cos}^{2} θ + 2}$
$2 = 4 {cos}^{2} θ$
$cos θ = \pm \frac{1}{\sqrt{2}}$
$θ = \frac{π}{4}$

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Similar questions

The eccentric angle $θ$ of a point $(\sqrt{6} cos θ, \sqrt{2} sin θ)$ on the ellipse $\frac{x^{2}}{6} + \frac{y^{2}}{2} = 1$ whose distance from the centre of the ellipse is $2$ is

Q. The area of the parallelogram formed by the tangents at the points whose eccentric angles

θ, θ + \frac{π}{2}, θ + π, θ + \frac{3 π}{2}

on the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

is:

Q. The distance of the point

(\sqrt{6} cos θ, \sqrt{2} sin θ)

on the ellipse

\frac{x^{2}}{6} + \frac{y^{2}}{2} = 1

from the centre is

2,

if:

Q. Assertion :The value of

θ, 0 < θ < \frac{π}{2},

satisfying the equation

6 \sum m = 1 c o s e c [θ + \frac{(m - 1) π}{4}] c o s e c [θ + \frac{m π}{4}] = 4 \sqrt{2}

are

\frac{π}{12}

and

\frac{5 π}{12}

Reason: If

tan θ + cot θ = 4, 0 < θ < \frac{π}{2},

then

θ = \frac{π}{12}

\frac{5 π}{12}

Q. If

\frac{π}{2} < θ < \frac{3 π}{2}

, then the value of

= \sqrt{4 {cos}^{4} θ + {sin}^{2} 2 θ} + 4 cot θ {cos}^{2} (\frac{π}{4} - \frac{θ}{2})

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The distance of a point (√6cosθ,√2sinθ) on the ellipse x26+y22=1 from the centre is 2, if:

The correct option is D θ=π4P(√6cosθ,√2sinθ)Centre(0,0)PC=√6cos2θ+2sin2θ2=√4cos2θ+22=4cos2θcosθ=±1√2θ=π4

The distance of a point $(\sqrt{6} cos θ, \sqrt{2} sin θ)$ on the ellipse $\frac{x^{2}}{6} + \frac{y^{2}}{2} = 1$ from the centre is $2$ , if:

The correct option is D $θ = \frac{π}{4}$
$P (\sqrt{6} cos θ, \sqrt{2} sin θ)$
$C e n t r e (0, 0)$
$P C = \sqrt{6 {cos}^{2} θ + 2 {sin}^{2} θ}$
$2 = \sqrt{4 {cos}^{2} θ + 2}$
$2 = 4 {cos}^{2} θ$
$cos θ = \pm \frac{1}{\sqrt{2}}$
$θ = \frac{π}{4}$