Ellipse MCQ Questions & Answers in Geometry | Maths

Let the ellipse be $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$
It passes through $$\left( { - 3,\,1} \right)$$  so $$\frac{9}{{{a^2}}} + \frac{1}{{{b^2}}} = 1.....({\text{i}})$$
Also, $${b^2} = {a^2}\left( {1 - \frac{2}{5}} \right)$$
$$ \Rightarrow 5{b^2} = 3{a^2}.....({\text{ii}})$$
Solving (i) and (ii) we get $${a^2} = \frac{{32}}{3},\,\,\,{b^2} = \frac{{32}}{5}$$
So, the equation of the ellipse is $$3{x^2} + 5{y^2} = 32$$

The ellipse can be written as, $$\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{16}} = 1$$
Here $${a^2} = 25,\,{b^2} = 16,$$
$$\eqalign{ & {\text{But }}{b^2} = {a^2}\left( {1 - {e^2}} \right) \cr & \Rightarrow \frac{{16}}{{25}} = 1 - {e^2} \cr & \Rightarrow {e^2} = 1 - \frac{{16}}{{25}} = \frac{9}{{25}} \cr & \Rightarrow e = \frac{3}{5} \cr} $$
Foci of the ellipse are $$\left( { \pm ae,\,0} \right) = \left( { \pm 3,\,0} \right),$$     i.e., $${F_1}$$ and $${F_2}$$
$$\therefore $$  We have $$P{F_1} + P{F_2} = 2a = 10$$     for every point $$P$$ on the ellipse.

Any point on the ellipse $$ = \left( {\sqrt 6 \cos \,\phi ,\,\sqrt 2 \sin \,\phi } \right).$$     So, $${\left( {\sqrt 6 \cos \,\phi - 0} \right)^2} + {\left( {\sqrt 2 \sin \,\phi - 0} \right)^2} = 4\,\,\, \Rightarrow \cos \,\phi = \pm \frac{1}{{\sqrt 2 }}$$

The equation of chord of contact of tangents from two points $$\left( {\alpha ,\,\beta } \right)$$  and $$\left( {\gamma ,\,\delta } \right)$$  to the given ellipse are
$$\eqalign{ & \frac{{x\alpha }}{5} + \frac{{y\beta }}{2} = 1......\left( 1 \right){\text{ and }}\frac{{x\gamma }}{5} + \frac{{y\delta }}{2} = 1......\left( 2 \right) \cr & {\text{Since}}\left( 1 \right){\text{and}}\left( 2 \right){\text{are}} \bot {\text{,}} \cr & \therefore \,\frac{{ - 2\alpha }}{{5\beta }} \times \frac{{ - 2\gamma }}{{5\delta }} = - 1\, \Rightarrow \frac{{\alpha \gamma }}{{\beta \delta }} = - \frac{{25}}{4} \cr} $$

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

$$\eqalign{ & {\text{Also }}{e^2} = 1 - \frac{{{b^2}}}{{{a^2}}} = 1 - {e^2} \cr & \Rightarrow 2{e^2} = 1,\,\,e = \frac{1}{{\sqrt 2 }} \cr} $$

The sum of distances of $$P$$ from the foci $$ = 2a = 2 \times 5 = 10$$

$$\eqalign{ & 2ae = 6\,\,\, \Rightarrow ae = 3;\,\,\,\,2b = 8\,\,\, \Rightarrow b = 4 \cr & {b^2} = {a^2}\left( {1 - {e^2}} \right); \cr & 16 = {a^2} - {a^2}{e^2} \cr & \Rightarrow {a^2} = 16 + 9 = 25 \cr & \Rightarrow a = 5 \cr & \therefore e = \frac{3}{a} = \frac{3}{5} \cr} $$

From standard equation of ellipse $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,$$    the co-ordinates of foci are

$$S\left( {ae,\,0} \right){\text{ and }}S'\left( { - ae,\,0} \right)$$
Co-ordinate of an extremity of the minor axis is $$B\left( {0,\,b} \right)$$
Now slope of straight line $$BS = \frac{{b - 0}}{{0 - ae}} = \frac{b}{{ - ae}} = {m_1}$$
and slope of straight line $$BS' = \frac{{b - 0}}{{0 - \left( { - ae} \right)}} = \frac{b}{{ae}} = {m_2}$$
$$\eqalign{ & \because \,BS \bot BS',\,{\text{so }}{m_1}.{m_2} = - 1 \cr & {\text{or }}\frac{b}{{ - ae}} \times \frac{b}{{ae}} = - 1\,; \cr & {\text{or }}\,\frac{{{b^2}}}{{{a^2}}} = {e^2} \cr & {\text{But }}{b^2} = {a^2}\left( {1 - {e^2}} \right)\,; \cr & {\text{or }}\,\frac{{{b^2}}}{{{a^2}}} = 1 - {e^2}\,; \cr & {\text{or }}\,{e^2} = 1 - {e^2}; \cr & {\text{or }}\,2{e^2} = 1\,; \cr & {\text{or }}\,e = \frac{1}{{\sqrt 2 }} \cr} $$

We know that two diameters $$y = {m_1}x,\,y = {m_2}x$$     are conjugate diameters of $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$   if $${m_1}{m_2} = - \frac{{{b^2}}}{{{a^2}}}$$
Hence, we have $$1.\left( { - \frac{2}{3}} \right) = - \frac{{{b^2}}}{{{a^2}}}{\text{ or }}{b^2} = \frac{2}{3}{a^2}$$
$$\therefore \,\,{b^2} = {a^2}\left( {1 - {e^2}} \right)\,\, \Rightarrow \frac{2}{3}{a^2} = {a^2}\left( {1 - {e^2}} \right)\,\, \Rightarrow \frac{2}{3} = 1 - {e^2}$$

The given ellipse is $$\frac{{{x^2}}}{9} + \frac{{{y^2}}}{5} = 1$$
Then $${a^2} = 9,\,\,{b^2} = 5\,\, \Rightarrow e = \sqrt {1 - \frac{5}{9}} = \frac{2}{3}$$
$$\therefore $$ end point of latus rectum in first quadrant is $$L\left( {2,\,\frac{5}{3}} \right)$$
Equation of tangent at $$L$$ is $$\frac{{2x}}{9} + \frac{y}{3} = 1$$
It meets $$x$$-axis at $$A\left( {\frac{9}{2},\,0} \right)$$   and $$y$$-axis at $$B\left( {0,\,3} \right)$$
$$\therefore $$ Area of $$\Delta OAB = \frac{1}{2} \times \frac{9}{2} \times 3 = \frac{{27}}{4}$$

By symmetry area of quadrilateral
$$ = 4 \times \left( {{\text{Area of }}\Delta OAB} \right) = 4 \times \frac{{27}}{4} = 27{\text{ sq}}{\text{. units}}$$