Question
The slope of the diameter of the ellipse $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,$$ whose length is the GM of the major and minor axes, is :
A.
$$\sqrt {\frac{a}{b}} $$
B.
$$\sqrt {ab} $$
C.
$$\sqrt {\frac{b}{a}} $$
D.
$$\frac{a}{b}$$
Answer :
$$\sqrt {\frac{b}{a}} $$
Solution :
The length of the diameter $$ = \sqrt {2a.2b} = 2\sqrt {ab} $$
Let $$P\left( {a\cos \phi ,\,b\sin \,\phi } \right)$$ be one end of the diameter.
Then $$2\left\{ {{{\left( {a\cos \phi - 0} \right)}^2} + {{\left( {b\sin \,\phi - 0} \right)}^2}} \right\} = 2\sqrt {ab} $$
or $${a^2}{\cos ^2}\phi + {b^2}{\sin ^2}\phi = ab$$
The slope of the diameter $$ = \frac{{b\sin \,\phi - 0}}{{a\cos \,\phi - 0}} = \frac{b}{a}\tan \,\phi $$
Now, $${a^2} + {b^2}{\tan ^2}\phi = ab{\sec ^2}\phi = ab\left( {1 + {{\tan }^2}\phi } \right)$$
$$\eqalign{
& \therefore \,\,\,{\tan ^2}\phi = \frac{{ab - {a^2}}}{{{b^2} - ab}} = \frac{{a\left( {b - a} \right)}}{{b\left( {b - a} \right)}} = \frac{a}{b} \cr
& \Rightarrow \frac{{{b^2}}}{{{a^2}}}{\tan ^2}\phi = \frac{b}{a} \cr
& \therefore \,\,{\left( {{\text{slope}}} \right)^2} = \frac{b}{a} \cr} $$