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Byju's Answer

Standard XII

Mathematics

Properties of Argument

Let S and ...

Question

Let $S$ and $S^{'}$ be the focii of an ellipse. At any point $P$ on the ellipse if $S P S^{'} < 90^{0}$ then the eccentricity is:

e > \frac{1}{\sqrt{2}}

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e < \frac{1}{\sqrt{2}}

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e = \frac{1}{\sqrt{2}}

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e = \frac{1}{2}

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Solution

The correct option is B $e < \frac{1}{\sqrt{2}}$

${(2 a e)}^{2} = a^{2} + a^{2}$
$4 a^{2} e^{2} = {2 a}^{2}$
$e^{2} = \frac{1}{2}$
$e = \frac{1}{\sqrt{2}}$
At $θ < 90^{0}$ , e will keep decreasing then
$(e < \frac{1}{\sqrt{2}})$

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Similar questions

Q. Let

S_{1}, S_{2}

are foci of an ellipse, whose major axis length is

15

units and

P

be any point on the ellipse such that perimeter of triangle

P S_{1} S_{2}

20

units. If

e

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3 e =

Q. If

S

and

S^{'}

be the foci of the ellipse

(x^{2} / a^{2}) + (y^{2} / b^{2}) = 1

and

e

be its eccentricity, then

tan \frac{P S S^{'}}{4} tan \frac{P S^{'} S}{4} = \frac{1 - e}{1 + e}

where

P

is any point on the ellipse.

The tangent at any point P on the ellipse meets the auxiliary

circle in two points which subtend right angle at the centre of the ellipse. If θ be the eccentric angle of P and e be the eccentricity of the ellipse, then sinθ =

Q. Let

S, S^{'}

be the foci of the ellipse

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

whose eccentricity is '

e

P

is a variable point on the ellipse. Consider the locus of the incentre of the

△ P S S^{'}

The eccentricity of locus of

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Q. If

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with foci

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then area of

Δ S P S^{'}

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Let S and S′ be the focii of an ellipse. At any point P on the ellipse if SPS′<900 then the eccentricity is:

The correct option is B e<1√2(2ae)2=a2+a24a2e2=2a2e2=12e=1√2At θ<900, e will keep decreasing then(e<1√2)

Let $S$ and $S^{'}$ be the focii of an ellipse. At any point $P$ on the ellipse if $S P S^{'} < 90^{0}$ then the eccentricity is:

The correct option is B $e < \frac{1}{\sqrt{2}}$

${(2 a e)}^{2} = a^{2} + a^{2}$
$4 a^{2} e^{2} = {2 a}^{2}$
$e^{2} = \frac{1}{2}$
$e = \frac{1}{\sqrt{2}}$
At $θ < 90^{0}$ , e will keep decreasing then
$(e < \frac{1}{\sqrt{2}})$