Question

The points on the \[4{x^2} + 9{y^2} = 13\] such that the rate of decrease of ordinate is equal to the rate of increase in abscissa are-
\[({\text{A}})\;\;\left( {\dfrac{3}{2},\dfrac{2}{3}} \right)\] \[({\text{B}})\;\;\left( {\dfrac{{ - 3}}{2},\dfrac{{ - 2}}{3}} \right)\]
\[({\text{C}})\;\;\left( {\dfrac{3}{2},\dfrac{{ - 2}}{3}} \right)\] \[({\text{D}})\;\;\left( {\dfrac{{ - 3}}{2},\dfrac{2}{3}} \right)\]

Answer 1

Hint: Here, we will be differentiating the given equation with respect to time ‘t’ as mentioned in the question, as rate of decrease in ordinate is equal to the increase in abscissa, we would find the values of ‘x’ or ‘y’ and will substitute it in the given equation to get the points.

Complete step-by-step answer:
The equation of the ellipse mentioned is \[4{x^2} + 9{y^2} = 13\].
We all know that the equation of ellipse is,
\[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\], centered at the origin axes lying along the Coordinate axes. In general, an ellipse may be centered at any points, So have axes not parallel to the coordinate axes.
Ellipse:-
Equation:- \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\] where \[a \geqslant b;\]
And ‘a’ represents half the length of the major axis while ‘b’ represents half the length of minor axis.
Taking the eqn of ellipse,
\[4{x^2} + 9{y^2} = 13\;\;\;\;({\text{i}})\]
Differentiating (i) w.r.t. ‘t’ time,
\[4{x^2} + 9{y^2} = 13\]
\[
   = \dfrac{d}{{dt}}(4{x^2}) + \dfrac{d}{{dt}}(9{y^2}) = \dfrac{d}{{dt}}(13) \\
   = (4\; \times \;2)\;x\dfrac{{dx}}{{dt}} + (9\; \times \;2)y\dfrac{{dy}}{{dt}} = 0 \\
   = 8x\dfrac{{dx}}{{dt}} + 18y\dfrac{{dy}}{{dt}} = 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;({\text{ii}}) \\
\]
According to question,
Decrease in ordinate\[ = - \dfrac{{dx}}{{dt}}\],
Increase in abscissa \[ = \dfrac{{dy}}{{dt}}\]
Both are equal, So,
\[ - \dfrac{{dx}}{{dt}} = \dfrac{{dy}}{{dt}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;({\text{iii}})\]
Putting the value of eq. (iii) to eq. (ii)
\[
  8x\dfrac{{dx}}{{dt}} + 18y\dfrac{{dy}}{{dt}} = 0 \\
   = 8x\dfrac{{dx}}{{dt}} + 18y\left( { - \dfrac{{dx}}{{dt}}} \right) = 0 \\
   = \dfrac{{dx}}{{dt}}\;(8x - 18y) = 0 \\
   = 8x - 18y = 0 \\
   = 8x = 18y \\
   = 4x = 9y \\
   = x = \dfrac{9}{4}y \\
\]
Now, Substituting \[x = \dfrac{9}{4}y\;\;\;{\text{in eq}}{\text{. (i)}}\], we
Get,
\[
  4{x^2} + 9{y^2} = 13 \\
   = 4\left( {\dfrac{9}{4}} \right){y^2} + 9{y^2} = 13 \\
   = {\dfrac{9}{4}^2}{y^2} + 9{y^2} = 13 \\
   = \left( {\dfrac{{81}}{4} + 9} \right){y^2} = 13 \\
   = \left( {\dfrac{{81 + 36}}{4}} \right){y^2} = 13 \\
   = {y^2} = \dfrac{{13\; \times \;4}}{{81 + 36}} = \dfrac{{{{{{52}}}^4}}}{{{{{{117}}}^9}}} = \dfrac{4}{9} \\
   = y = \dfrac{2}{3} \\
  x = \dfrac{9}{4}\; \times \;x = \dfrac{{{{{9}}^3}}}{{{{{4}}^2}}}\; \times \;\dfrac{{{2}}}{{{3}}} = \dfrac{3}{2}. \\
\]
Coordinates : \[\left( {\dfrac{3}{2},\dfrac{2}{3}} \right)\;({\text{A}})\]

Note: While solving questions where we need to find the coordinate and question of the ellipse is given, we need to differentiate the given equation & substitute the given condition in the equation to get the desired relation between the x & y coordinates.