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Question

The point on the ellipse 16x² + 9y² = 400 where ordinate decrease at the same rate at which the abscissa increases is

(9, 16)
$(3, \frac{16}{3})$
$(\frac{16}{3}, 3)$
$(- 3 \frac{16}{3})$

hint Hint:

We are given the equation of ellipse. We are given that the rate of decrease of ordinante is same as the rate of increase of abscissa. Rate means change w.r.t the time. To find the rate of change of quantity, we take it's derivative w.r.t time. We have to find the point at which the rate is as given above.

The correct answer is: $open parentheses 3 comma fraction numerator 16 over denominator 3 end fraction close parentheses$

The given equation of ellipse is 16x² + 9y² = 400
Ordinate means distance of the point from x-axis. It is given by y coordinate.
Abscissa means distance of the point from the y-axis. It is given by x coordinate.
We are given that, $negative fraction numerator d y over denominator d t end fraction equals space fraction numerator d x over denominator d t end fraction$ .....(1)
We will take the derivate of the equation of ellipse w.r.t time.
$16 x squared space plus space 9 y squared space equals space 400 16 left parenthesis 2 right parenthesis x space fraction numerator d x over denominator d t end fraction plus space 9 left parenthesis 2 right parenthesis y fraction numerator d y over denominator d t end fraction space equals space 0 32 x space fraction numerator d x over denominator d t end fraction plus space 18 y fraction numerator d y over denominator d t end fraction equals 0 32 x fraction numerator d x over denominator d t end fraction equals space minus 18 y fraction numerator d y over denominator d t end fraction N o w comma space u sin g space t h e space e q u a t i o n space left parenthesis 1 right parenthesis space i n space t h e space a b o v e space e q u a t i o n 32 x space open parentheses negative fraction numerator d y over denominator d t end fraction close parentheses space equals 18 y open parentheses negative fraction numerator d y over denominator d t end fraction close parentheses 32 x space equals space 18 y D i v i d i n g space b o t h space t h e space s i d e s space b y space 2 space w e space g e t comma 16 x space equals 9 y space space space space x space equals 9 over 16 y space space$
Now, to find the points we will substitute the value of x in the equation of ellipse.
$16 x squared space plus space 9 y squared space equals space 400 16 open parentheses 9 over 16 close parentheses squared y squared space plus space 9 y squared space equals space 400 space space space space 9 squared over 16 y squared space plus space 9 y squared space equals space 400 space space space space space T a k i n g space 9 y squared space c o m m o n space 9 y squared open parentheses 9 over 16 space plus space 1 close parentheses space equals space 400 9 y squared open parentheses 25 over 16 close parentheses space equals space 400 M u l t i p l y i n g space b o t h space t h e space s i d e s space b y space open parentheses 16 over 25 close parentheses 9 y squared space equals space 400 space cross times space 16 over 25 9 y squared space equals space 16 space cross times space 16 y squared space equals space 256 over 9$
Taking square roots of both sides we get,
$y space equals plus-or-minus 16 over 3$
$x space equals space 9 over 16 y F o r space y space equals space plus 16 over 3 x space equals space 3 F o r space y space equals space minus 16 over 3 x space equals space minus 3$
So, the points satisfying the given condition are $open parentheses 3 comma 16 over 3 close parentheses space a n d space open parentheses negative 3 comma negative 16 over 3 close parentheses$
This is the only point that matches the option $open parentheses 3 comma 16 over 3 close parentheses$ .

.

For such questions, we should know how to find the derivates. We should know the concept of abscissa and ordinate.

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An YDSE is carried out in a liquid of refractive index $mu equals 1.3$ and a thin film of air is formed in front of the lower slit as shown in the figure. If a maxima of third order is formed at the origin O, find the thickness of the air film. The wavelength of light in air is $lambda subscript 0 end subscript equals 0.78 mu m$ and D/d=1000.

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Two coherent narrow slits emitting light of wavelength $lambda$ in the same phase are placed parallel to each other at a small separation of $2 lambda$ . The light is collected on a screen S which is placed at a distance $D left parenthesis much greater-than lambda right parenthesis$ from the slit $S subscript 1 end subscript$ as shown in figure. Find the finite distance x such that the intensity of P is equal to intensity at O.

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Consider the arrangement shown in figure. By some mechanism, the separation between the slits $S subscript 3 end subscript$ and $S subscript 4 end subscript$ can be changed. The intensity is measured at the point P which is at the common perpendicular bisector of $S subscript 1 end subscript S subscript 2 end subscript$ and $S subscript 3 end subscript S subscript 4 end subscript$ . When $z equals fraction numerator D lambda over denominator 2 blank d end fraction$ , the intensity measured at P is I. Find this intensity when z is equal to

1) $fraction numerator D lambda over denominator d end fraction$ ,
2) $fraction numerator 3 D lambda over denominator 2 blank d end fraction$ , and
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$stack l i m with x rightwards arrow pi divided by 2 below fraction numerator a to the power of cot invisible function application x end exponent minus a to the power of cos invisible function application x end exponent over denominator c o t invisible function application x minus c o s invisible function application x end fraction equals$

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In Young's double slit experiment $S subscript 1 end subscript$ and $S subscript 2 end subscript$ are two slits. Films of thicknesses $t subscript 1 end subscript$ and $t subscript 2 end subscript$ and refractive indices $m subscript 1 end subscript$ and $m subscript 2 end subscript$ are placed in front of $S subscript 1 end subscript$ and $S subscript 2 end subscript$ respectively. If $m subscript 1 end subscript t subscript 1 end subscript equals m subscript 2 end subscript t subscript 2 end subscript$ , then the central maximum will :

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Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities in the resulting beam are

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Can’t find what you’re looking for?

The point on the ellipse 16x² + 9y² = 400 where ordinate decrease at the same rate at which the abscissa increases is

(9, 16)
$(3, \frac{16}{3})$
$(\frac{16}{3}, 3)$
$(- 3 \frac{16}{3})$

The correct answer is: $open parentheses 3 comma fraction numerator 16 over denominator 3 end fraction close parentheses$

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The point on the ellipse 16x2 + 9y2 = 400 where ordinate decrease at the same rate at which the abscissa increases is

(9, 16)

The correct answer is:

Related Questions to study

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If and then is equal to

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A parallel beam of light 500 nm is incident at an angle 30° with the normal to the slit plane in a young's double slit experiment. The intensity due to each slit is Io. Point O is equidistant from and . The distance between slits is 1 mm.

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The point on the ellipse 16x² + 9y² = 400 where ordinate decrease at the same rate at which the abscissa increases is

(9, 16)
$(3, \frac{16}{3})$
$(\frac{16}{3}, 3)$
$(- 3 \frac{16}{3})$

The correct answer is: $open parentheses 3 comma fraction numerator 16 over denominator 3 end fraction close parentheses$

The displacement of a particle in time ‘t’ is given by S = t³ – t² – 8t – 18. The acceleration of the particle when its velocity vanishes is

The displacement of a particle in time ‘t’ is given by S = t³ – t² – 8t – 18. The acceleration of the particle when its velocity vanishes is

The figure shows a transparent slab of length 1 m placed in air whose refractive index in x direction varies as $mu equals 1 plus x to the power of 2 end exponent left parenthesis 0 less than x less than 1 right parenthesis$ . The optical path length of ray R will be

The figure shows a transparent slab of length 1 m placed in air whose refractive index in x direction varies as $mu equals 1 plus x to the power of 2 end exponent left parenthesis 0 less than x less than 1 right parenthesis$ . The optical path length of ray R will be

A parallel beam of light 500 nm is incident at an angle 30° with the normal to the slit plane in a young's double slit experiment. The intensity due to each slit is Io. Point O is equidistant from $S subscript 1 end subscript$ and $S subscript 2 end subscript$ . The distance between slits is 1 mm.

A parallel beam of light 500 nm is incident at an angle 30° with the normal to the slit plane in a young's double slit experiment. The intensity due to each slit is Io. Point O is equidistant from $S subscript 1 end subscript$ and $S subscript 2 end subscript$ . The distance between slits is 1 mm.