Ellipse MCQ Questions & Answers in Geometry | Maths

$$\left( {\alpha ,\,\beta } \right)$$  is an exterior point if $$16{\alpha ^2} + 9{\beta ^2} - 16\alpha - 32 > 0.$$

The given ellipse is $$\frac{{{x^2}}}{{{4^2}}} + \frac{{{y^2}}}{{{2^2}}} = 1$$
such that $${a^2} = 16$$   and $${b^2} = 4$$
$$\therefore {e^2} = 1 - \frac{4}{{16}} = \frac{3}{4}\,\, \Rightarrow e = \frac{{\sqrt 3 }}{2}$$
Let $$P\left( {4\,\cos \,\theta ,\,2\,\sin \,\theta } \right)$$    be any point on the ellipse, then equation of normal at $$P$$ is
$$\eqalign{ & 4x\,\sin \,\theta - 2y\,\cos \,\theta = 12\,\sin \,\theta \,\cos \,\theta \cr & \Rightarrow \frac{x}{{3\,\cos \,\theta }} - \frac{y}{{6\,\sin \,\theta }} = 1 \cr} $$
$$\therefore \,\,Q,$$ the point where normal at $$P$$ meets x-axis, has coordinates $$\left( {3\,\cos \,\theta ,\,0} \right)$$
$$\therefore $$ Mid point of $$PQ$$  is $$M\left( {\frac{{7\,\cos \,\theta }}{2},\,\sin \,\theta } \right)$$
For locus of point $$M$$ we consider
$$\eqalign{ & x = \frac{{7\,\cos \,\theta }}{2}{\text{ and }}y = \sin \,\theta \cr & \Rightarrow \cos \,\theta = \frac{{2x}}{7}{\text{ and }}\sin \,\theta = y \cr & \Rightarrow \frac{{4{x^2}}}{{49}} + {y^2} = 1.....(1) \cr} $$
Also the latus rectum of given ellipse is
$$\eqalign{ & x = \pm ae = \pm 4 \times \frac{{\sqrt 3 }}{2} = \pm 2\sqrt 3 \cr & {\text{or }}\,\,x = \pm 2\sqrt 3 .....(2) \cr} $$
Solving equations (1) and (2), we get
$$\eqalign{ & \frac{{4 \times 12}}{{49}} + {y^2} = 1 \cr & \Rightarrow {y^2} = \frac{1}{{49}}{\text{ or }}y = \pm \frac{1}{7} \cr} $$
$$\therefore $$ The required points are $$\left( { \pm 2\sqrt 3 ,\, \pm \frac{1}{7}} \right)$$

Given: $${x^2} - {y^2}{\sec ^2}\theta = 4$$     and $${x^2}{\sec ^2}\theta + {y^2} = 16$$
$$ \Rightarrow \frac{{{x^2}}}{4} - \frac{{{y^2}}}{{4\,{{\cos }^2}\theta }} = 1{\text{ and }}\frac{{{x^2}}}{{16\,{{\cos }^2}\theta }} + \frac{{{y^2}}}{{16}} = 1$$
According to problem
$$\eqalign{ & \frac{{4 + 4\,{{\cos }^2}\theta }}{4} = 3\left( {\frac{{16 - 16\,{{\cos }^2}\theta }}{{16}}} \right) \cr & \Rightarrow 1 + {\cos ^2}\theta = 3\left( {1 - {{\cos }^2}\theta } \right) \cr & \Rightarrow 4\,{\cos ^2}\theta = 2 \cr & \Rightarrow \cos \,\theta = \pm \frac{1}{2} \cr & \Rightarrow \theta = \frac{\pi }{4},\,\frac{{3\pi }}{4} \cr} $$

Given equation of ellipse is $$\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{16}} = 1$$
$$ \Rightarrow a = 5{\text{ and }}b = 4$$
Since $$p\left( {x,\,y} \right)$$  moves along ellipse and $$C$$ is the center
$$\eqalign{ & \therefore \,\max \left( {CP} \right) = 5{\text{ and }}\min \left( {CP} \right) = 4 \cr & \therefore 4\,\max \left\{ {cp} \right\} + 5\,\min \left\{ {cp} \right\} = 4 \times 5 + 5 \times 4 = 40 \cr} $$

Let the equation of the ellipse be $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$
Let $$e$$ be the eccentricity of the ellipse.
Since distance between foci $$ = 2h$$
$$\therefore \,2ae = 2h \Rightarrow e = h......\left( 1 \right)$$
Focal distance of one end of minor axis say $$\left( {0,\,b} \right)$$  is $$k$$
$$\therefore \,a + e\left( 0 \right) = k \Rightarrow a = k......\left( 2 \right)$$
From $$\left( 1 \right)$$ and $$\left( 2 \right),\,\,{b^2} = {a^2}\left( {1 - {e^2}} \right) = {k^2} - {h^2}$$
$$\therefore $$  The equation of the ellipse is $$\frac{{{x^2}}}{{{k^2}}} + \frac{{{y^2}}}{{{k^2} - {h^2}}} = 1.$$

For ellipse $$\frac{{{x^2}}}{{{4^2}}} + \frac{{{y^2}}}{{{3^2}}} = 1,\,\,a = 4,\,\,b = 3$$
$$ \Rightarrow e = \sqrt {1 - {{\left( {\frac{3}{4}} \right)}^2}} = \frac{{\sqrt 7 }}{4}$$
$$\therefore $$ Foci are $$\left( {\sqrt 7 ,0} \right)$$   and $$\left( { - \sqrt 7 ,0} \right)$$
Centre of circle is at (0, 3) and it passes through $$\left( { + \sqrt 7 ,0} \right),$$   therefore radius of circle $$ = \sqrt {{{\left( {\sqrt 7 } \right)}^2} + {{\left( 3 \right)}^2}} = 4$$

The end point of latus rectum of ellipse $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$   in first quadrant is $$\left( {ae,\,\frac{{{b^2}}}{a}} \right)$$  and the tangent at this point intersects $$x$$-axis at $$\left( {\frac{a}{e},\,0} \right)$$  and $$y$$-axis at $$\left( {0,\,a} \right)$$
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
$$ = 4 \times \left( {{\text{Area of }}\Delta OAB} \right) = 4 \times \frac{{27}}{4} = 27{\text{ sq}}{\text{. units}}$$

Let the point $$P$$ be $$\left( {a\,\cos \,\theta ,\,b\,\sin \,\theta } \right)$$
The equation of tangent at $$P$$ is

$$\frac{{x\,\cos \,\theta }}{a} + \frac{{y\,\sin \,\theta }}{b} = 1......\left( 1 \right)$$
If $$d$$ be the length of perpendicular from the centre $$C\left( {0,\,0} \right)$$  of the ellipse to the tangent given by $$\left( 1 \right)$$ then
$$\eqalign{ & d = \frac{1}{{\sqrt {\frac{{{{\cos }^2}\theta }}{{{a^2}}} + \frac{{{{\sin }^2}\theta }}{{{b^2}}}} }} \cr & \Rightarrow \frac{1}{{{d^2}}} = \frac{{{{\cos }^2}\theta }}{{{a^2}}} + \frac{{{{\sin }^2}\theta }}{{{b^2}}} \cr & \Rightarrow \frac{{{b^2}}}{{{d^2}}} = \frac{{{b^2}}}{{{a^2}}}{\cos ^2}\theta + 1 - {\cos ^2}\theta \cr & \Rightarrow 1 - \frac{{{b^2}}}{{{d^2}}} = \left( {1 - \frac{{{b^2}}}{{{a^2}}}} \right){\cos ^2}\theta \cr & \Rightarrow 1 - \frac{{{b^2}}}{{{d^2}}} = {e^2}{\cos ^2}\theta ......\left( 2 \right) \cr & {\text{Now,}}\,\,{\left( {P{F_1} - P{F_2}} \right)^2} \cr & = {\left( {2ae\,\cos \,\theta } \right)^2} \cr & = 4{a^2}{e^2}{\cos ^2}\theta \cr & = 4{a^2}\left( {1 - \frac{{{b^2}}}{{{d^2}}}} \right) \cr} $$

The distance between foci $$ = 2ae = \sqrt {{{\left( {3 - 1} \right)}^2} + {{\left( {1 - 1} \right)}^2}} = 2\,\,\,\,\,\,\therefore \,ae = 1$$
The sum of the distances of foci from $$\left( {1,\,3} \right)$$  on the ellipse $$ = 2a$$
$$\eqalign{ & \Rightarrow \sqrt {{{\left( {3 - 1} \right)}^2} + {{\left( {1 - 3} \right)}^2}} + \sqrt {{{\left( {1 - 1} \right)}^2} + {{\left( {1 - 3} \right)}^2}} = 2a \cr & \Rightarrow \,2a = 2\sqrt 2 + 2 \cr & \therefore \,e = \frac{1}{a} = \frac{2}{{2\sqrt 2 + 2}} = \frac{1}{{\sqrt 2 + 1}} = \sqrt 2 - 1 \cr} $$

Let $$\left( {{x_1},\,{y_1}} \right)$$  be their point of intersection then
$$\eqalign{ & x_1^2 - y_1^2 = {c^2}......\left( 1 \right) \cr & \frac{{x_1^2}}{{{a^2}}} + \frac{{y_1^2}}{{{b^2}}} = 1......\left( 2 \right) \cr & \Rightarrow x_1^2\left( {\frac{1}{{{a^2}}} - \frac{1}{{{c^2}}}} \right) + y_1^2\left( {\frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}} \right) = 0......\left( 3 \right) \cr} $$
Now tangents to the curves are $$x{x_1} - y{y_1} = {c^2}$$    and $$\frac{{x{x_1}}}{{{a^2}}} + \frac{{y{y_1}}}{{{b^2}}} = 1$$
The tangents are perpendicular, so
$$\frac{{{x_1}}}{{{y_1}}} \times - \frac{{{b^2}}}{{{a^2}}}\frac{{{x_1}}}{{{y_1}}} = - 1 \Rightarrow {b^2}x_1^2 - {a^2}y_1^2 = 0......\left( 4 \right)$$
Eliminating $$x_1^2$$ and $$y_1^2$$ from $$\left( 3 \right)$$ and $$\left( 4 \right),$$ we get,
$$\frac{{{c^2} - {a^2}}}{{{a^2}{b^2}{c^2}}} = \frac{{ - \left( {{b^2} + {c^2}} \right)}}{{{a^2}{b^2}{c^2}}} \Rightarrow {a^2} - {b^2} = 2{c^2}$$