Question
A point on the ellipse $$\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1$$ at a distance equal to the mean of the lengths of the semi-major axis and semi-minor axis from the centre is :
A.
$$\left( {\frac{{2\sqrt {91} }}{7},\,\frac{{3\sqrt {105} }}{{14}}} \right)$$
B.
$$\left( {\frac{{2\sqrt {91} }}{7},\, - \frac{{3\sqrt {105} }}{{14}}} \right)$$
C.
$$\left( {\frac{{2\sqrt {105} }}{7},\,\frac{{3\sqrt {91} }}{{14}}} \right)$$
D.
$$\left( { - \frac{{2\sqrt {105} }}{7}, - \,\frac{{3\sqrt {91} }}{{14}}} \right)$$
Answer :
$$\left( {\frac{{2\sqrt {91} }}{7},\,\frac{{3\sqrt {105} }}{{14}}} \right)$$
Solution :
Let the point is $$\left( {4\,\cos \,\theta ,\,3\,\sin \,\theta } \right)$$
According to question,
$$\eqalign{
& {\left( {4\,\cos \,\theta } \right)^2} + {\left( {3\,\sin \,\theta } \right)^2} = {\left( {\frac{{4 + 3}}{2}} \right)^2}......\left( 1 \right) \cr
& {\text{From equation }}\left( 1 \right), \cr
& 16 - 7\,{\sin ^2}\theta = \frac{{49}}{4}\,\, \Rightarrow {\sin ^2}\theta = \frac{{15}}{{28}} \cr
& \therefore \,\sin \,\theta = \pm \frac{1}{2}\sqrt {\frac{{15}}{7}} = \pm \frac{{\sqrt {105} }}{{14}} \cr} $$
Similarly, $$\cos \,\theta = \pm \frac{{\sqrt {91} }}{{14}}$$
So the points are $$\left( {\frac{{2\sqrt {91} }}{7},\,\frac{{3\sqrt {105} }}{{14}}} \right)\,;\,\left( { - \frac{{2\sqrt {91} }}{7},\, - \frac{{3\sqrt {105} }}{{14}}} \right)$$
Interchange $$\theta $$ by $$\frac{\pi }{2} + \theta $$ and $$\frac{{3\pi }}{2} + \theta $$