Ellipse MCQ Questions & Answers in Geometry | Maths

Since $${1^2} + {2^2} = 5 < 9$$    and $${2^2} + {1^1} = 5 < 9$$    both $$P$$ and $$Q$$ lie inside $$C.$$
Also $$\frac{{{1^2}}}{9} + \frac{{{2^2}}}{4} = \frac{1}{9} + 1 > 1$$     and $$\frac{{{2^2}}}{9} + \frac{1}{4} = \frac{{25}}{{36}} < 1,\,\,P$$     lies outside $$E$$ and $$Q$$ lies inside $$E.$$  Thus $$P$$ lies inside $$C$$ but outside $$E.$$

Given equation of ellipse is
$$\eqalign{ & \frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} = 1 \cr & \Rightarrow \frac{{{x^2}}}{{{{\left( 2 \right)}^2}}} + \frac{{{y^2}}}{{{{\left( 3 \right)}^2}}} = 1 \cr & \Rightarrow a = 2{\text{ and }}b = 3 \cr} $$
Length of major axis $$ = 2a = 4$$
Since, we have
Sum of the focal distances of a point on ellipse $$=$$ length of major axis.
$$\therefore $$  Required Answer $$ = 4$$  units.

$$\eqalign{ & 2b = 2ae,\,\,\frac{{2{b^2}}}{a} = 10,\,\,{b^2} = {a^2}\left( {1 - {e^2}} \right) \cr & \Rightarrow e = \frac{1}{{\sqrt 2 }},\,\,a = 10,\,{b^2} = 50 \cr} $$

Let the point $$P$$ be $$\left( {a\,\cos \,\theta ,\,b\,\sin \,\theta } \right)$$
The tangent at $$P$$ is $$\frac{x}{a}\cos \,\theta + \frac{y}{b}\sin \,\theta = 1......\left( {\text{i}} \right)$$
The perpendicular distance $$p$$ of $$S\left( {ae,\,0} \right)$$   form $$\left( {\text{i}} \right)$$ is given by
$$\eqalign{ & {p^2} = \frac{{{{\left( {e\,\cos \,\theta - 1} \right)}^2}}}{{\frac{{{{\cos }^2}\theta }}{{{a^2}}} + \frac{{{{\sin }^2}\theta }}{{{b^2}}}}} \cr & \Rightarrow \frac{1}{{{p^2}}} = \frac{{\frac{{{{\cos }^2}\theta }}{{{a^2}}} + \frac{{{{\sin }^2}\theta }}{{{b^2}}}}}{{{{\left( {e\,\cos \,\theta - 1} \right)}^2}}} \cr & \Rightarrow \frac{{{b^2}}}{{{p^2}}} = \frac{{\frac{{{b^2}}}{{{a^2}}}{{\cos }^2}\theta + 1 - {{\cos }^2}\theta }}{{{{\left( {e\,\cos \,\theta - 1} \right)}^2}}} \cr & \Rightarrow \frac{{{b^2}}}{{{p^2}}} = \frac{{\left( {\frac{{{b^2}}}{{{a^2}}} - 1} \right){{\cos }^2}\theta + 1}}{{{{\left( {e\,\cos \,\theta - 1} \right)}^2}}} \cr & \Rightarrow \frac{{{b^2}}}{{{p^2}}} = \frac{{1 - {e^2}{{\cos }^2}\theta }}{{{{\left( {e\,\cos \,\theta - 1} \right)}^2}}} \cr & \Rightarrow \frac{{{b^2}}}{{{p^2}}} = \frac{{1 + e\,\cos \,\theta }}{{1 - e\,\cos \,\theta }} \cr & {\text{Now, }}SP = a\left( {1 - e\,\cos \,\theta } \right) \cr & \therefore \,\frac{{2a}}{{SP}} - 1 = \frac{{2a}}{{a\left( {1 - e\,\cos \,\theta } \right)}} - 1 = \frac{{1 + e\,\cos \,\theta }}{{1 - e\,\cos \,\theta }} = \frac{{{b^2}}}{{{p^2}}} \cr} $$

$$\frac{{{x^2}}}{9} + \frac{{{y^2}}}{3} = 1\,\,\,\, \Rightarrow {a^2} = 9,\,\,{b^2} = 3$$
So, $${b^2} = {a^2}\left( {1 - {e^2}} \right)\,\, \Rightarrow 3 = 9\left( {1 - {e^2}} \right)\,\, \Rightarrow e = \sqrt {\frac{2}{3}} $$
$$\therefore $$  distance between foci $$ = 2ae = 2.3\sqrt {\frac{2}{3}} $$

$$\eqalign{ & {\text{Here, }}{a^2} = 9,\,\,{b^2} = 5 \cr & {\text{So, latus rectum}} = \frac{{2{b^2}}}{a} = \frac{{2 \times 5}}{3} \cr & \therefore \,P = \left( {\alpha ,\,\frac{5}{3}} \right){\text{.}}\,{\text{It is on the ellipse}}{\text{.}} \cr & {\text{So, }}\frac{{{\alpha ^2}}}{9} + \frac{{{{\left( {\frac{5}{3}} \right)}^2}}}{5} = 1{\text{ or }}\alpha = \pm 2.\,\,\,\,{\text{So, }}P = \left( {2,\,\frac{5}{3}} \right) \cr & {\text{The tangent at }}P{\text{ is }}\frac{{2x}}{9} + \frac{{\frac{5}{3}y}}{5} = 1{\text{ or }}\frac{x}{{\frac{9}{2}}} + \frac{y}{3} = 1 \cr & \therefore \,\,{\text{area}} = 4 \times {\text{ar}}\left( {\Delta ROQ} \right) = 4 \times \left( {\frac{1}{2} \times \frac{9}{2} \times 3} \right) = 27 \cr} $$

$$\eqalign{ & \because \,\angle FBF' = {90^ \circ } \Rightarrow F{B^2} + F'{B^2} = FF{'^2} \cr & \therefore {\left( {\sqrt {{a^2}{e^2} + {b^2}} } \right)^2} + {\left( {\sqrt {{a^2}{e^2} + {b^2}} } \right)^2} = {\left( {2ae} \right)^2} \cr & \Rightarrow 2\left( {{a^2}{e^2} + {b^2}} \right) = 4{a^2}{e^2} \cr & \Rightarrow {e^2} = \frac{{{b^2}}}{{{a^2}}}......\left( {\text{i}} \right) \cr} $$

$$\eqalign{ & {\text{Also, }}{e^2} = 1 - \frac{{{b^2}}}{{{a^2}}} = 1 - {e^2}\,\,\,\left( {{\text{By using equation}}\left( {\text{i}} \right)} \right) \cr & \Rightarrow 2{e^2} = 1 \cr & \Rightarrow e = \frac{1}{{\sqrt 2 }} \cr} $$

Let the ellipse be $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1......\left( 1 \right)$$
Then $${e^2} = 1 - \frac{{{b^2}}}{{{a^2}}}......\left( 2 \right)$$
Let a point $$P$$ on $$\left( 1 \right)$$ be $$\left( {a\,\cos \,\theta ,\,b\,\sin \,\theta } \right)$$
The coordinates of foci are $${S_1}\left( {ae,\,0} \right)$$   and $${S_2}\left( { - ae,\,0} \right)$$
Hence, $${S_1}P = a\left( {1 - e\,\cos \,\theta } \right)$$
$${S_2}P = a\left( {1 + e\,\cos \,\theta } \right)$$     and $${S_1}{S_2} = 2ae$$
If $$\left( {h,\,k} \right)$$  be the coordinates of in centre then
$$\eqalign{ & h = \frac{{2ae \times a\,\cos \,\theta + a\left( {1 - e\,\cos \,\theta } \right) \times - ae + a\left( {1 + e\,\cos \,\theta } \right) \times ae}}{{2ae + a\left( {1 - e\,\cos \,\theta } \right) + a\left( {1 + e\,\cos \,\theta } \right)}} \cr & = \frac{{2ae\,\cos \,\theta }}{{1 + e}}......\left( 3 \right) \cr & k = \frac{{be\,\sin \,\theta }}{{1 + e}}......\left( 4 \right) \cr} $$
Squaring and adding $$\left( 3 \right)$$ & $$\left( 4 \right)$$ we have,
$$\frac{{{h^2}}}{{4{a^2}}} + \frac{{{k^2}}}{{{b^2}}} = {\left( {\frac{e}{{1 + e}}} \right)^2}$$
$$\therefore $$  The locus of the point $$\left( {h,\,k} \right)$$  is
$$\frac{{{x^2}}}{{4{a^2}{\lambda ^2}}} + \frac{{{y^2}}}{{{b^2}{\lambda ^2}}} = 1,{\text{ where }}\lambda = \frac{e}{{1 + e}}$$
Which is another ellipse with eccentricity $$ = \sqrt {1 - \frac{{{b^2}}}{{4{a^2}}}} = \sqrt {\frac{{3 + {e^2}}}{4}} $$

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, $${a^2} + {b^2}{\tan ^2}\phi = ab{\sec ^2}\phi = ab\left( {1 + {{\tan }^2}\phi } \right)$$
$$\eqalign{ & \therefore \,\,\,{\tan ^2}\phi = \frac{{ab - {a^2}}}{{{b^2} - ab}} = \frac{{a\left( {b - a} \right)}}{{b\left( {b - a} \right)}} = \frac{a}{b} \cr & \Rightarrow \frac{{{b^2}}}{{{a^2}}}{\tan ^2}\phi = \frac{b}{a} \cr & \therefore \,\,{\left( {{\text{slope}}} \right)^2} = \frac{b}{a} \cr} $$

Clearly $$P$$ is $$\left( {a\,\cos \,\theta ,\,b\,\sin \,\theta } \right)$$   and $$Q$$ is $$\left( { - a\,\sin \,\theta ,\,b\,\sin \,\theta } \right)$$    so the mid point $$\left( {h,\,k} \right)$$  of $$PQ$$  will be given by
$$\eqalign{ & h = \frac{{a\,\cos \,\theta - a\,\sin \,\theta }}{2}{\text{ and }}k = \frac{{b\,\sin \,\theta + b\,\cos \,\theta }}{2} \cr & \therefore \,\frac{{4{h^2}}}{{{a^2}}} + \frac{{4{k^2}}}{{{b^2}}} = 2 \Rightarrow \frac{{{h^2}}}{{{a^2}}} + \frac{{{k^2}}}{{{b^2}}} = \frac{1}{2} \cr} $$