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### Question

#### Tangent at any point of ellipse is drawn. Eccentric

angle of ‘ P’ is If is the foot of perpendicular from centre to this tangent then is

#### The correct answer is:

_{}

_{}

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### Related Questions to study

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#### If the curve subtends as obtuse angle at the point , then a possible value of is

The generated curve is , whose director circle is . For the required condition should lie inside the circle and out side the ellipse i.e.

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The generated curve is , whose director circle is . For the required condition should lie inside the circle and out side the ellipse i.e.

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#### Let . A point *P* in the XY-plane varies in such a way that perimeter of is 16. Then the maximum area of is

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#### From the focus of the ellipse a ray of light is sent which makes angle with the positive direction of X-axis upon reacting the ellipse the ray is reflected from it. Slope of the reflected ray is

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Foci are

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This line meets the ellipse above X-axis at

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#### From the focus of the ellipse a ray of light is sent which makes angle with the positive direction of X-axis upon reacting the ellipse the ray is reflected from it. Slope of the reflected ray is

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Equation of line through (-5, 0) with slope –2 is

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#### If the tangent at Point P to the ellipse 16x^{2} + 11y^{2} = 256 is also the tangent to the circle x^{2} + y^{2} - 2x = 15, then the eccentric angle of point P is

The equation of tangent at point to the ellipse

16x

16x (4cosq) + 11y = 256

4xcosq + y sinq = 16

This touches the circle

(x + 1)

So, = 4

Þ (cosq - 4)

4cos

\ cosq =

\ q = ±

16x

^{2}+ 11y^{2}= 256 is16x (4cosq) + 11y = 256

4xcosq + y sinq = 16

This touches the circle

(x + 1)

^{2}+ y^{2}= 16So, = 4

Þ (cosq - 4)

^{2}= 11 + 5cos^{2}q4cos

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botany-General

The equation of tangent at point to the ellipse

16x

16x (4cosq) + 11y = 256

4xcosq + y sinq = 16

This touches the circle

(x + 1)

So, = 4

Þ (cosq - 4)

4cos

\ cosq =

\ q = ±

16x

^{2}+ 11y^{2}= 256 is16x (4cosq) + 11y = 256

4xcosq + y sinq = 16

This touches the circle

(x + 1)

^{2}+ y^{2}= 16So, = 4

Þ (cosq - 4)

^{2}= 11 + 5cos^{2}q4cos

^{2}q + 8cosq - 5 = 0\ cosq =

\ q = ±

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#### Determine the potential at point B, assuming point A to be at zero potential

<img src="https://mycourses.turito.com/tokenpluginfile.php/c161933dbfaab094c54655ab71e9b8f0/1/question/questiontext/869608/1/201040/Picture9.jpeg" alt="" width="160" height="108"

From the loop BDCR

2 ´ 2 + 3R

R

\ V

_{1}B, we get2 ´ 2 + 3R

_{1}= 4R

_{1}= 0\ V

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Physics-General

From the loop BDCR

2 ´ 2 + 3R

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\ V

_{1}B, we get2 ´ 2 + 3R

_{1}= 4R

_{1}= 0\ V

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