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Maths-

General

Easy

Question

If P( $theta$ ), Q $open parentheses theta plus fraction numerator pi over denominator 2 end fraction close parentheses$ are points on ellipse and $alpha$ is angle between normals at P and Q then -

$2 \sqrt{1 - e^{2}}$ = ${e s i n}^{2} 2 θ t a n α$
$2 \sqrt{1 - e^{2}}$ = $e {s i n}^{2} θ \cdot t a n 2 α$
$\sqrt{1 - e^{2}}$ = $2 e^{2}$ ${s i n}^{2} 2 θ \cdot t a n α$
$2 \sqrt{1 - e^{2}}$ = $e^{2} {s i n}^{2} 2 θ \cdot t a n α$

The correct answer is: $2 square root of 1 – e to the power of 2 end exponent end root$ = $e to the power of 2 end exponent s i n to the power of 2 end exponent invisible function application 2 theta times t a n invisible function application alpha$

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The Maxwell's four equations are written as:
i) $not stretchy contour integral stack E with rightwards arrow on top. stack d s with rightwards arrow on top equals fraction numerator q subscript 0 end subscript over denominator epsilon subscript 0 end subscript end fraction$
ii) $not stretchy contour integral stack B with rightwards arrow on top. stack d s with rightwards arrow on top equals 0$
iii) $not stretchy contour integral stack E with rightwards arrow on top. stack d l with rightwards arrow on top equals fraction numerator d over denominator d t end fraction not stretchy contour integral stack B with rightwards arrow on top. stack d s with rightwards arrow on top$
iv) $not stretchy contour integral stack B with rightwards arrow on top. stack d s with rightwards arrow on top equals mu subscript 0 end subscript epsilon subscript 0 end subscript fraction numerator d over denominator d t end fraction not stretchy contour integral stack E with rightwards arrow on top. stack d s with rightwards arrow on top$
The equations which have sources of $stack E with rightwards arrow on top$ and $stack B with rightwards arrow on top$

The Maxwell's four equations are written as:
i) $not stretchy contour integral stack E with rightwards arrow on top. stack d s with rightwards arrow on top equals fraction numerator q subscript 0 end subscript over denominator epsilon subscript 0 end subscript end fraction$
ii) $not stretchy contour integral stack B with rightwards arrow on top. stack d s with rightwards arrow on top equals 0$
iii) $not stretchy contour integral stack E with rightwards arrow on top. stack d l with rightwards arrow on top equals fraction numerator d over denominator d t end fraction not stretchy contour integral stack B with rightwards arrow on top. stack d s with rightwards arrow on top$
iv) $not stretchy contour integral stack B with rightwards arrow on top. stack d s with rightwards arrow on top equals mu subscript 0 end subscript epsilon subscript 0 end subscript fraction numerator d over denominator d t end fraction not stretchy contour integral stack E with rightwards arrow on top. stack d s with rightwards arrow on top$
The equations which have sources of $stack E with rightwards arrow on top$ and $stack B with rightwards arrow on top$

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