Ellipse MCQ Questions & Answers in Geometry | Maths

For ellipse, $$ab - {h^2} > 0\,\,\, \Rightarrow \left( {k + 1} \right).1 - {\left( {k - 1} \right)^2} > 0{\text{ or }}k\left( {k - 3} \right) < 0$$

Equation of ellipse is $$\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{8} = 1$$    where, $$a = 4,\,b = 2\sqrt 2 $$
Eccentricity, $$e = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} = \sqrt {1 - \frac{8}{{16}}} = \frac{1}{{\sqrt 2 }}$$
Area is maximum when vertex is $$\left( {0,\,b} \right)$$
$$\therefore $$  Maximum area
$$\eqalign{ & = \frac{1}{2} \times 2ae \times b \cr & = \frac{1}{2} \times 2 \times 4 \times 2\sqrt 2 \times \frac{1}{{\sqrt 2 }} \cr & = 8{\text{ square units}} \cr} $$

Tangent at $$\left( {{x_1},\,{y_1}} \right)$$  is $$16{x_1}x + 9{y_1}y = 144.$$    Intercepts on axes are equal $$ \Rightarrow \frac{{144}}{{16{x_1}}} = \frac{{144}}{{9{y_1}}}{\text{ or }}9{y_1} = 16{x_1}{\text{. Also, }}16x_1^2 + 9y_1^2 = 144$$
Solving these, $${x_1} = \frac{9}{5},\, - \frac{9}{5}{\text{ and }}{y_1} = \frac{{16}}{5},\, - \frac{{16}}{5}$$
So, the tangent is $$16 \times \frac{9}{5}x + 9 \times \frac{{16}}{5}y = 144{\text{ or }}16 \times \frac{{ - 9}}{5}x + 9 \times \frac{{ - 16}}{5}y = 144$$
$$ \Rightarrow \,\,x + y = 5{\text{ or }}x + y + 5 = 0$$

From the given equation of ellipse, we have
$$\eqalign{ & a = 4,\,b = 3,\,e = \sqrt {1 - \frac{9}{{16}}} \cr & \Rightarrow e = \frac{{\sqrt 7 }}{4} \cr} $$

Now, radius of this circle $$ = {a^2} = 16$$
$$ \Rightarrow $$ Foci $$ = \left( { \pm \sqrt 7 ,\,0} \right)$$
Now equation of circle is $${\left( {x - 0} \right)^2} + {\left( {y - 3} \right)^2} = 16$$
$${x^2} + {y^2} - 6y - 7 = 0$$

Length of latus rectum of an ellipse is $$\frac{{2{b^2}}}{a}$$  where $$b$$ is semi minor axis and $$a$$ is semimajor axis. As given,
$$\frac{{2{b^2}}}{a} = b\, \Rightarrow 2b = a\, \Rightarrow \frac{b}{a} = \frac{1}{2}$$
We know that eccentricity $$e = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} = \sqrt {1 - \frac{1}{4}} = \frac{{\sqrt 3 }}{2}$$

The given ellipse is $${x^2} + 9{y^2} = 9$$   or $$\frac{{{x^2}}}{{{3^2}}} + \frac{{{y^2}}}{{{1^2}}} = 1$$

So, that $$A\left( {3,\,0} \right)$$   and $$B\left( {0,\,1} \right)$$
$$\therefore $$ Equation of $$AB$$  is $$\frac{x}{3} + \frac{y}{1} = 1$$
or $$x + 3y - 3 = 0.....(1)$$
Also auxiliary circle of given ellipse is
$${x^2} + {y^2} = 9.....(2)$$
Solving equation (1) and (2), we get the point $$M$$ where line $$AB$$  meets the auxiliary circle.
Putting $$x=3-3y$$   from equation (1) in equation (2)
we get $${\left( {3 - 3y} \right)^2} + {y^2} = 9$$
$$\eqalign{ & \Rightarrow 9 - 18y + 9{y^2} + {y^2} = 9 \cr & \Rightarrow 10{y^2} - 18y = 0 \cr & \Rightarrow y = 0,\,\frac{9}{5}\,\,\,\,\,\,\, \Rightarrow x = 3,\,\frac{{ - 12}}{5} \cr} $$
Clearly $$M\left( {\frac{{ - 12}}{5},\,\frac{9}{5}} \right)$$
$$\therefore $$ Area of \[\Delta OAM = \frac{1}{2}\left| \begin{array}{l} \,\,\,0\,\,\,\,\,\,\,\,\,\,0\,\,\,\,1\\ \,\,\,3\,\,\,\,\,\,\,\,\,\,0\,\,\,\,1\\ \frac{{ - 12}}{5}\,\,\,\frac{9}{5}\,\,\,1 \end{array} \right| = \frac{{27}}{{10}}\]

For centre, $$\frac{{\partial S}}{{\partial x}} = 0,\,\frac{{\partial S}}{{\partial y}} = 0,\,\,\, \Rightarrow 7x - y - 11 = 0,\,\,2x - 11y + 29 = 0.$$           Solve.

$$\eqalign{ & {\text{The ellipse is }}\frac{{{x^2}}}{{{3^2}}} + \frac{{{y^2}}}{{{2^2}}} = 1. \cr & {\text{So, }}P = \left( {3\cos \,\phi ,\,2\sin \,\phi } \right){\text{ and }}Q = \left( {3\cos \,\overline {\phi + \frac{\pi }{2}} ,\,2\sin \,\overline {\phi + \frac{\pi }{2}} } \right) \cr} $$
\[\begin{array}{l} \therefore \,\,{\rm{area \,of\, the\, }}\Delta CPQ = \frac{1}{2}\left| \begin{array}{l} \,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\ \,\,\,\,\,\,3\cos \,\phi \,\,\,\,\,2\sin \,\phi \,\,\,\,\,\,\,\,\,\,\,\,1\\ - 3\sin \,\phi \,\,\,\,\,2\cos \,\phi \,\,\,\,\,\,\,\,\,\,\,\,1\, \end{array} \right|\\ = \frac{1}{2}\left( {6{{\cos }^2}\phi + 6{{\sin }^2}\phi } \right)\\ = \frac{1}{2}.6\\ = 3\, \end{array}\]

Latus rectum $$ = \frac{{2{b^2}}}{a} = \frac{2}{a}.{a^2}\left( {1 - {e^2}} \right).$$

Centre is $$\left( {1,\,2} \right)$$  and focus is $$\left( {6,\,2} \right),$$  hence the line joining the centre $$C$$ and the focus $$S$$ (i.e., the axis) is parallel to $$x$$-axis. Therefore, the equation must be of the form

$$\frac{{{{\left( {x - 1} \right)}^2}}}{{{a^2}}} + \frac{{{{\left( {y - 2} \right)}^2}}}{{{b^2}}} = 1......\left( {\text{i}} \right)$$
The distance between the centre and the focus
$$CS = ae = 6 - 1 = 5$$
Also
$${b^2} = {a^2}\left( {1 - {e^2}} \right) = {a^2} - {a^2}{e^2} = {a^2} - 25$$
$$\therefore $$  The equation $$\left( {\text{i}} \right)$$ becomes
$$\frac{{{{\left( {x - 1} \right)}^2}}}{{{a^2}}} + \frac{{{{\left( {y - 2} \right)}^2}}}{{{a^2} - 25}} = 1......\left( {{\text{ii}}} \right)$$
The point $$\left( {4,\,6} \right)$$  lies on it, therefore
$$\eqalign{ & \frac{{{{\left( {4 - 1} \right)}^2}}}{{{a^2}}} + \frac{{{{\left( {6 - 2} \right)}^2}}}{{{a^2} - 25}} = 1\, \Rightarrow {a^2} = 45 \cr & \therefore \,{b^2} = 45 - 25 = 20 \cr} $$
The required equation is $$\frac{{{{\left( {x - 1} \right)}^2}}}{{45}} + \frac{{{{\left( {y - 2} \right)}^2}}}{{20}} = 1$$

$$\eqalign{ & 9ac - 9{a^2} - 2{c^2} > 9ac - 6{c^2}......\left( {\text{i}} \right) \cr & {\text{Again }}3a < 2c \cr & \Rightarrow 9ac < 6{c^2} \cr & \Rightarrow 9ac - 6{c^2}......\left( {{\text{ii}}} \right) \cr & {\text{From }}\left( {\text{i}} \right){\text{and}}\left( {{\text{ii}}} \right),\,\,9ac - 9{a^2} - 2{c^2} > 0 \cr} $$