Ellipse MCQ Questions & Answers in Geometry | Maths

Equation of the given conic is an equation of ellipse $$\frac{{{x^2}}}{{{a^2} + \lambda }} + \frac{{{y^2}}}{{{b^2} + \lambda }}\,\left( {x \geqslant 0} \right)$$
$$ \Rightarrow {A^2} = {a^2} + \lambda {\text{ and }}{B^2} = {b^2} + \lambda $$
$$\eqalign{ & {\text{Eccentricity, }}e = \sqrt {1 - \frac{{{B^2}}}{{{A^2}}}} \cr & = \sqrt {1 - \frac{{{b^2} + \lambda }}{{{a^2} + \lambda }}} \cr & = \sqrt {\frac{{{a^2} + \lambda - {b^2} - \lambda }}{{{a^2} + \lambda }}} \cr & = \sqrt {\frac{{{a^2} - {b^2}}}{{{a^2} + \lambda }}} \cr} $$
$$\lambda $$ is in the denominator so, when $$\lambda $$ increases, the eccentricity decreases.

Let $$P = \left( {\sqrt 2 a\cos \,\phi ,\,\sqrt 2 b\sin \,\phi } \right).\,{F_1}{\text{ and }}{F_2} = \left( { \pm \sqrt 2 ae,\,0} \right)$$
\[\begin{array}{l} {\rm{The \,area\, of\, }}\Delta PFF' = \left| {\frac{1}{2}\left| \begin{array}{l} \sqrt 2 a\cos \,\phi \,\,\,\,\,\sqrt 2 b\sin \,\phi \,\,\,\,\,\,\,\,\,\,\,\,1\\ \,\,\,\,\,\,\,\,\sqrt 2 ae\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\ \,\,\, - \sqrt 2 ae\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \end{array} \right|} \right|\\ = \frac{1}{2}.\sqrt 2 b\sin \,\phi .\sqrt 2 ae\\ = 2abe\sin \,\phi \end{array}\]
$$\therefore $$  maximum area $$ = 2abc = 2ab.\frac{{\sqrt {{a^2} + {b^2}} }}{a}$$

Given equation of ellipse $$E$$ is
$$\eqalign{ & \frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1 \cr & \Rightarrow \frac{{4{x^2} + 9{y^2}}}{{36}} = 1 \cr & \Rightarrow 4{x^2} + 9{y^2} = 36 \cr & \Rightarrow 4{x^2} + 9{y^2} - 36 = 0......\left( 1 \right) \cr} $$
And $$C\,:$$  equation of circle is $${x^2} + {y^2} = 9$$
Which can be rewritten as $${x^2} + {y^2} - 9 = 0......\left( 2 \right)$$
For a point $$P = \left( {1,\,2} \right)$$   we have
$$\eqalign{ & 4{\left( 1 \right)^2} + 9{\left( 2 \right)^2} - 36 = 40 - 36 > 0\,\,\,\,\left[ {{\text{from equation }}\left( 1 \right)} \right] \cr & \,{\text{and }}\,{1^2} + {2^2} - 9 = 5 - 9 < 0\,\,\,\,\,\left[ {{\text{from equation }}\left( 2 \right)} \right] \cr} $$
$$\therefore $$  Point $$P$$ lies outside of $$E$$ and inside of $$C.$$

The given ellipse is $$\frac{{{x^2}}}{4} + \frac{{{y^2}}}{1} = 1$$
So $$A = \left( {2,\,0} \right)$$   and $$B = \left( {0,\,1} \right)$$
If $$PQRS$$   is the rectangle in which it is inscribed, then $$P = \left( {2,\,1} \right).$$
Let $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$
be the ellipse circumscribing the rectangle $$PQRS.$$
Then it passes through $$P\left( {2,\,1} \right)$$

$$\therefore \frac{4}{{{a^2}}} + \frac{1}{{{b^2}}} = 1.....(a)$$
Also, given that, it passes through $$\left( {4,\,0} \right)$$
$$\eqalign{ & \therefore \frac{{16}}{{{a^2}}} + 0 = 1\,\,\,\, \Rightarrow {a^2} = 16 \cr & \Rightarrow {b^2} = \frac{4}{3}\left[ {{\text{substituting }}{a^2} = 16{\text{ in equation }}\left( a \right)} \right] \cr} $$
$$\therefore $$ The required ellipse is $$\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{\frac{4}{3}}} = 1{\text{ or }}{x^2} + 12{y^2} = 16$$

Any tangent to ellipse $$\frac{{{x^2}}}{2} + \frac{{{y^2}}}{1} = 1$$   is
$$\frac{{x\,\cos \,\theta }}{{\sqrt 2 }} + y\,\sin \,\theta = 1$$

$$\eqalign{ & \therefore A\left( {\sqrt 2 \,\sec \,\theta ,\,0} \right)\,;\,B\left( {0,\,{\text{cosec}}\,\theta } \right) \cr & \Rightarrow 2h = \sqrt 2 \,\sec \,\theta \,\,{\text{and}}\,\,2k = {\text{cosec}}\,\theta \left( {{\text{using mid pt}}{\text{. formula}}} \right) \cr & \Rightarrow \cos \,\theta = \frac{1}{{\sqrt 2 \,h}}{\text{ and }}\sin \,\theta = \frac{1}{{2k}} \cr & \Rightarrow {\left( {\frac{1}{{\sqrt 2 \,h}}} \right)^2} + {\left( {\frac{1}{{2k}}} \right)^2} = 1 \cr & \Rightarrow \frac{1}{{2{h^2}}} + \frac{1}{{4{k^2}}} = 1 \cr} $$
Required locus,
$$\frac{1}{{2{x^2}}} + \frac{1}{{4{y^2}}} = 1$$

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 $$\frac{{3\pi }}{2} + \theta $$

Any tangent to the ellipse $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$    at $$P\left( {a\,\cos \,\theta ,\,b\,\sin \,\theta } \right)$$    is $$\frac{{x\,\cos \,\theta }}{a} + \frac{{y\,\sin \,\theta }}{b} = 1$$

It meets co-ordinate axes at $$A\left( {a\,\sec \,\theta ,\,0} \right)$$   and $$B\left( {0,\,b\,{\text{cosec}}\,\theta } \right)$$
$$\therefore $$ Area of $$\Delta OAB = \frac{1}{2} \times a\,\sec \,\theta \times b\,{\text{cosec}}\,\theta $$
$$ \Rightarrow \Delta = \frac{{ab}}{{\sin \,2\theta }}$$
For $$\Delta $$ to be min, $${\sin \,2\theta }$$  should be max. and we know max. value of $$\sin \,2\theta = 1$$
$$\therefore {\Delta _{\max }} = ab\,\,{\text{sq}}{\text{. units}}$$

Equation of the ellipse is $$\frac{{{x^2}}}{9} + \frac{{{y^2}}}{1} = 1$$
An end of the major axis $$A$$ be say $$\left( {3,\,0} \right)$$  and an end of the minor axis $$B$$ be say $$\left( {0,\,1} \right).$$  Equation of $$AB$$  is therefore.
$$\frac{x}{3} + \frac{y}{1} = 1......\left( 1 \right)$$

Equation of the auxiliary circle is $${x^2} + {y^2} = 9......\left( 2 \right)$$
Solving the equation $$\left( 1 \right)$$ and $$\left( 2 \right)$$ we get
$$\eqalign{ & {x^2} + {\left( {1 - \frac{x}{3}} \right)^2} = 9 \cr & \Rightarrow {x^2} + 1 + \frac{{{x^2}}}{9} - \frac{{2x}}{3} = 9 \cr & \Rightarrow 5{x^2} - 3x - 36 = 0 \cr & \Rightarrow \left( {5x + 12} \right)\left( {x - 3} \right) = 0 \cr & \therefore \,x = - \frac{{12}}{5} \Rightarrow y = 1 - \frac{1}{3}\left( { - \frac{{12}}{5}} \right) = \frac{9}{5} \cr & \therefore {\text{ Coordinates of }}M{\text{ are }}\left( { - \frac{{12}}{5},\,\frac{9}{5}} \right) \cr & {\text{Area of }}\Delta AOM = \frac{1}{2}.OA.MN = \frac{1}{2} \times 3 \times \frac{9}{5} = \frac{{27}}{{10}} \cr} $$

$$c = \sqrt {{a^2}{m^2} + {b^2}} {\text{ where }}a = 1,\,\,b = 4,\,\,m = 3.$$

Here, $$9{x^2} + 16{y^2} = 144$$     and $$3x + 4y = 12$$
$$ \Rightarrow x = \frac{{12 - 4y}}{3}$$
So, $$9{\left( {\frac{{12 - 4y}}{3}} \right)^2} + 16{y^2} = 144$$
On solving we get, $$y = 0,\,3$$
For $$y = 0\,;\,x = 4$$
For $$y = 3\,;\,x = 0$$
$$ \Rightarrow $$  Length of chord
$$\eqalign{ & = \sqrt {{{\left( {0 - 3} \right)}^2} + {{\left( {4 - 0} \right)}^2}} \cr & = \sqrt {9 + 16} \cr & = \sqrt {25} \cr & = 5{\text{ units}} \cr} $$