1.
Equation of the ellipse whose axes are the axes of coordinates and which passes through the point $$\left( { - 3,\,1} \right)$$ and has eccentricity $$\sqrt {\frac{2}{5}} $$ is :
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 $$3{x^2} + 5{y^2} = 32$$
2.
If $$P \equiv \left( {x,\,y} \right),\,{F_1} \equiv \left( {3,\,0} \right),\,{F_2} \equiv \left( { - 3,\,0} \right)$$ and $$16{x^2} + 25{y^2} = 400,$$ then $$P{F_1} + P{F_2}$$ equals :
Any point on the ellipse $$ = \left( {\sqrt 6 \cos \,\phi ,\,\sqrt 2 \sin \,\phi } \right).$$ So, $${\left( {\sqrt 6 \cos \,\phi - 0} \right)^2} + {\left( {\sqrt 2 \sin \,\phi - 0} \right)^2} = 4\,\,\, \Rightarrow \cos \,\phi = \pm \frac{1}{{\sqrt 2 }}$$
4.
If the chords of contact of tangents from two points $$\left( {\alpha ,\,\beta } \right)$$ and $$\left( {\gamma ,\,\delta } \right)$$ to the ellipse $$\frac{{{x^2}}}{5} + \frac{{{y^2}}}{2} = 1$$ are perpendicular, then $$\frac{{\alpha \gamma }}{{\beta \delta }} = \,?$$
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
$$\eqalign{
& \frac{{x\alpha }}{5} + \frac{{y\beta }}{2} = 1......\left( 1 \right){\text{ and }}\frac{{x\gamma }}{5} + \frac{{y\delta }}{2} = 1......\left( 2 \right) \cr
& {\text{Since}}\left( 1 \right){\text{and}}\left( 2 \right){\text{are}} \bot {\text{,}} \cr
& \therefore \,\frac{{ - 2\alpha }}{{5\beta }} \times \frac{{ - 2\gamma }}{{5\delta }} = - 1\, \Rightarrow \frac{{\alpha \gamma }}{{\beta \delta }} = - \frac{{25}}{4} \cr} $$
5.
An ellipse has $$OB$$ as semi minor axis, $$F$$ and $$F'$$ its focci and the angle $$FBF'$$ is a right angle. Then the eccentricity of the ellipse is :
6.
A point $$P$$ on the ellipse $$\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{9} = 1$$ has the eccentric angle $$\frac{\pi }{8}.$$ The sum of the distances of $$P$$ from the two foci is :
8.
If the angle between the straight lines joining the foci to an extremity of minor axis in an ellipse be $${90^ \circ },$$ then the eccentricity of the ellipse is :
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 $$BS' = \frac{{b - 0}}{{0 - \left( { - ae} \right)}} = \frac{b}{{ae}} = {m_2}$$
$$\eqalign{
& \because \,BS \bot BS',\,{\text{so }}{m_1}.{m_2} = - 1 \cr
& {\text{or }}\frac{b}{{ - ae}} \times \frac{b}{{ae}} = - 1\,; \cr
& {\text{or }}\,\frac{{{b^2}}}{{{a^2}}} = {e^2} \cr
& {\text{But }}{b^2} = {a^2}\left( {1 - {e^2}} \right)\,; \cr
& {\text{or }}\,\frac{{{b^2}}}{{{a^2}}} = 1 - {e^2}\,; \cr
& {\text{or }}\,{e^2} = 1 - {e^2}; \cr
& {\text{or }}\,2{e^2} = 1\,; \cr
& {\text{or }}\,e = \frac{1}{{\sqrt 2 }} \cr} $$
9.
If $$y = x$$ and $$3y + 2x = 0$$ are the equations of a pair of conjugate diameters of the ellipse $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$ then its eccentricity is :
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 $$1.\left( { - \frac{2}{3}} \right) = - \frac{{{b^2}}}{{{a^2}}}{\text{ or }}{b^2} = \frac{2}{3}{a^2}$$
$$\therefore \,\,{b^2} = {a^2}\left( {1 - {e^2}} \right)\,\, \Rightarrow \frac{2}{3}{a^2} = {a^2}\left( {1 - {e^2}} \right)\,\, \Rightarrow \frac{2}{3} = 1 - {e^2}$$
10.
The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse $$\frac{{{x^2}}}{9} + \frac{{{y^2}}}{5} = 1,$$ is-
A.
$$\frac{{27}}{4}\,\,{\text{sq}}{\text{.}}\,{\text{units}}$$
B.
$$9\,\,{\text{sq}}{\text{.}}\,{\text{units}}$$
C.
$$\frac{{27}}{2}\,\,{\text{sq}}{\text{.}}\,{\text{units}}$$
D.
$$27\,\,{\text{sq}}{\text{.}}\,{\text{units}}$$
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
$$ = 4 \times \left( {{\text{Area of }}\Delta OAB} \right) = 4 \times \frac{{27}}{4} = 27{\text{ sq}}{\text{. units}}$$