Conic Sections - Class 11 Commerce Maths - Extra Questions

Find the latus rectum, the eccentricity, and the coordinates of the foci, of the ellipses
(1) $$x^{2} + 3y^{2} = a^{2}$$, (2) $$5x^{2} + 4y^{2} = 1$$, and (3) $$9x^{2} + 5y^{2} - 30y = 0$$.

(1) $${ x }^{ 2 }+3{ y }^{ 2 }={ a }^{ 2 }$$ 

$$\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ \left( \dfrac { { a }^{ 2 } }{ 3 }  \right)  } =1$$

Latus rectum$$=\dfrac { 2{ b }^{ 2 } }{ a } =\dfrac { 2\times \dfrac { { a }^{ 2 } }{ 3 }  }{ { a } } =\dfrac { 2a }{ 3 } $$

$${ e }^{ 2 }=1-\dfrac { { b }^{ 2 } }{ { a }^{ 2 } } =1-\dfrac { \dfrac { { a }^{ 2 } }{ 3 }  }{ { a }^{ 2 } } =1-\dfrac { 1 }{ 3 } =\dfrac { 2 }{ 3 } \\ \Rightarrow e=\sqrt { \dfrac { 2 }{ 3 }  }=\dfrac { 1 }{ 3 } \sqrt { 6 }  $$

Coordinates of foci are $$(\pm ae,0)$$

$$\left( \pm \dfrac { a }{ 3 } \sqrt { 6 } ,0 \right) $$


(2) $$5{ x }^{ 2 }+4{ y }^{ 2 }=1$$ 

$$\dfrac { { x }^{ 2 } }{ { \left( \dfrac { 1 }{ \sqrt { 5 }  }  \right)  }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { \left( \dfrac { 1 }{ 2 }  \right)  }^{ 2 } } =1\\ a=\dfrac { 1 }{ \sqrt { 5 }  } ,b=\dfrac { 1 }{ 2 } \\ b>a$$

So the major axis of ellipse is $$y$$ axis

Latus rectum $$=\dfrac { 2{ a }^{ 2 } }{ b } =\dfrac { 2\times \dfrac { 1 }{ 5 }  }{ { \dfrac { 1 }{ 2 }  } } =\dfrac { 4 }{ 5 } $$

$${ e }^{ 2 }=1-\dfrac { { a }^{ 2 } }{ { b }^{ 2 } } =1-\dfrac { \dfrac {1}{5}}{\dfrac{1}{4} } =1-\dfrac { 4 }{ 5 } =\dfrac { 1 }{ 5 } \\ \Rightarrow e=\dfrac { 1 }{\sqrt{5}}$$

Foci : $$(0,\pm be)$$

$$\Rightarrow \left( 0,\pm \dfrac { 1 }{ 2\sqrt { 5 }  }  \right) $$


(3) $$9{ x }^{ 2 }+5{ y }^{ 2 }-30y=0$$ 

$$9{ x }^{ 2 }+5({ y }^{ 2 }-6y+9-9)=0\\ 9{ x }^{ 2 }+5{ (y-3) }^{ 2 }=45\\ \dfrac { { x }^{ 2 } }{ { 5 } } +\dfrac { { (y-3) }^{ 2 } }{ 9 } =1\\ a=\sqrt { 5 } ,b=3\\ b>a$$

So, the major axis of ellipse is $$y$$ axis.

Latus rectum $$=\dfrac { 2{ a }^{ 2 } }{ b } =\dfrac { 2\times 5 }{ { 3 } } =\dfrac { 10 }{ 3 } $$

$${ e }^{ 2 }=1-\dfrac { { a }^{ 2 } }{ { b }^{ 2 } } =1-\dfrac { 5 }{ 9 } =\dfrac { 4 }{ 9 } \\ \Rightarrow e=\dfrac { 2 }{ 3 } $$

For $$\dfrac { { (x-h) }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { (y-k) }^{ 2 } }{ { b }^{ 2 } } =1$$ with $$b>a$$

Foci is $$(h,\pm be+k)$$

Here $$h=0$$ and $$k=3$$

So foci are $$\left( 0,3\times \dfrac { 2 }{ 3 } +3 \right) =(0,5)$$ and $$\left( 0,-3\times \dfrac { 2 }{ 3 } +3 \right) =(0,1)$$

 



If the latus rectum of an ellipse is equal to half of minor axis, find its eccentricity.

$$\textbf{Step-1: Finding relation between a and b}$$
                $$\text{Latus rectum of ellipse is half of the minor axis}$$

                $$\Rightarrow \text{L.R} = \dfrac{{2{b^2}}}{a} = {b}$$

                $$ \Rightarrow \dfrac{b}{a} = \dfrac{1}{2}$$              $$\dots(1)$$

$$\textbf{Step-2: Eccentricity of ellipse is given as-}$$

                 $$e = \sqrt {1 - {{\left( {\dfrac{b}{a}} \right)}^2}} $$ $$\dots(2)$$

$$\textbf{Step-3: Putting values (1) in (2)}$$

                  $$e = \sqrt {1 - {{\left( {\dfrac{b}{a}} \right)}^2}}$$

                     $$ = \sqrt {1 - {{\left( {\dfrac{1}{2}} \right)}^2}}  $$

                     $$= \sqrt {1 - \dfrac{1}{{4}}} $$

                     $$ = \sqrt {\dfrac{{3}}{{4}}}$$

$$\textbf{So, the eccentricity of ellipse is given by}$$ $$ \mathbf{{\dfrac{{\sqrt3}}{{2}}}}. $$



If the eccentricity of an ellipse is $$\dfrac{5}{8}$$ and the distance between its foci is $$10$$, then find the latusrectum of the ellipse.

Let the equation of the required ellipse be $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$ and let $$e$$ be its eccentricity.

$$e=\dfrac{5}{8}$$ and $$2ae=10$$

$$e=\dfrac{5}{8}$$ and $$ae=5$$

$$e=\dfrac{5}{8}$$ and $$a=8$$

Therefore,
$$b^2=a^2(1-e^2)$$

$$b^2=64(1-\dfrac{25}{64})=39$$

Hence, length of the latusrectum = $$\dfrac{2b^2}{a}=2\times \dfrac{39}{8}=\dfrac{39}{4}$$


Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $$\displaystyle \frac{x^{2}}{49}+\frac{y^{2}}{36}= 1.$$

The given equation is $$\displaystyle \frac{x^{2}}{49}+\frac{y^{2}}{36}=

1$$ or $$\displaystyle \frac{x^{2}}{7^{2}}+\frac{y^{2}}{6^{2}}= 1$$
Her the donominator of $$\displaystyle \frac{x^{2}}{49} $$is greater than the donominator of $$\displaystyle \frac{y^{2}}{36}$$
Therefore the major axis is along the $$x$$-axis while the minor axis is along the $$y$$-axis
On comparing the given equation with $$\displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1$$ we obtain $$a = 7$$ and $$b = 6$$
$$\displaystyle \therefore ae=c=\sqrt{a^{2}-b^{2}}=\sqrt{49-36}=\sqrt{13}$$
Therefore, the coordinates of the foci are $$\displaystyle \left ( \pm \sqrt{13}, 0 \right )$$
The coordinates of the vertices are $$\displaystyle \left ( \pm 7, 0 \right )$$
Length of major axis $$= 2a = 14$$
Length of minor axis $$= 2b = 12$$
Eccentricity $$e =$$ $$\displaystyle \frac{c}{a}=$$ $$\displaystyle \frac{\sqrt{13}}{7}$$
Length of latus rectum $$=$$ $$\displaystyle \frac{2b^{2}}{a}=\frac{2\times 36}{7}=\frac{72}{7}$$


Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $$\displaystyle \frac{x^{2}}{100}+\frac{y^{2}}{400}=1.$$

The given equation is $$\displaystyle

\frac{x^{2}}{100}+\frac{y^{2}}{400}=1$$ or $$\displaystyle

\frac{x^{2}}{10^{2}}+\frac{y^{2}}{20^{2}}=1$$
Here the denominator of $$\displaystyle \frac{y^{2}}{400}$$ is greater than the denominator of $$\displaystyle \frac{x^{2}}{100}$$
Therefore, the major axis is alon the $$y$$-axis while the minor axis is along the $$x$$-axis.
On comparaing, the given equation with $$\displaystyle \frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}= 1$$ we obtain $$b = 10$$ and $$a = 20$$
$$\displaystyle \therefore ae=c = \sqrt{a^{2}-b^{2}}=\sqrt{400 - 100}= \sqrt{300}= 10\sqrt{3}$$
The coordinates of the foci are $$\displaystyle \left ( 0, \pm 10\sqrt{3} \right )$$
The coordinates of the vertices are $$\displaystyle \left ( 0, \pm  20 \right )$$
Length of major axis $$= 2a = 40$$
Length of minor axis $$= 2b = 20$$
Eccentricity $$e =$$ $$\displaystyle \frac{c}{a}=\frac{10\sqrt{3}}{20}= \frac{\sqrt{3}}{2}$$
Length of latus rectum $$=$$ $$\displaystyle \frac{2b^{2}}{a}= \frac{2\times 100}{20}= 10$$


Find the eccentricity of that ellipse, whose latus rectum is half of the minor axis.

Latus rectum $$=\dfrac{2b^{2}}{a} $$

Length of minor axis $$= 2b $$

Given:
Latus rectum $$= \dfrac{1}{2}$$(length of minor axis)

$$\therefore \dfrac{2b^{2}}{a} = \dfrac{1}{2}(2b) $$

$$\dfrac{b}{a}=\dfrac{1}{2}$$ 

Eccentricity $$e=\dfrac{c}{a}$$ where $$c=\sqrt{(a^{2}-b^{2}) }$$ 

$$ e=\sqrt{(1-\dfrac {b^{2}}{a^{2}}) } $$

$$=\sqrt{1-{\dfrac{1^2}{2^2}} }$$

$$e=\dfrac{\sqrt{3}}{2}$$



Find the equation of the parabola whose focus is $$(1, -1)$$ and vertex is $$(2, 1)$$

$$\textbf{Step-1: Find the value of a using distance formula.}$$
                 $$\text{Vertex form of parabola whose vertex is}$$ $$\left( h, k \right)$$.
                 $${\left( y - k \right)}^{2} = 4a \left( x - h \right)$$
                 $$\text{where as,}$$
                 $$a =$$ $$\text{distance between vertex and focus of parabola.}$$
                 $$\because$$ $$\text{vertex is given, i.e.,}$$ $$\left( 2, 1 \right)$$,
                 $$\therefore$$ $$\text{Equation of parabola is-}$$
                 $${\left( y - 1 \right)}^{2} = 4a \left( x - 2 \right)$$
                 $$\text{As the focus is given, i.e.,}$$ $$\left( 1, -1 \right)$$,
                 $$\therefore \; a = \sqrt{{\left( 2 - 1 \right)}^{2} + {\left( 1 - \left( -1 \right) \right)}^{2}}$$
                 $$\Rightarrow \; a = \sqrt{{1}^{2} + {2}^{2}}$$
                 $$\Rightarrow \; a = \sqrt{1 + 4} = \sqrt{5}$$
$$\textbf{Step-2: Use the above results to get the required unknown.}$$
                 $$\therefore$$ $$\text{Equation of parabola is-}$$
                 $${\left( y - 1 \right)}^{2} = 4 \times \sqrt{5} \times \left( x - 2 \right)$$
                 $${\left( y - 1 \right)}^{2} = 4 \sqrt{5} \left( x - 2 \right)$$
$$\textbf{Hence, answer.}$$


Find the latus rectum $$\&$$ equation of parabola whose vertex is origin $$\&$$ direct is $$x + y = 2.$$

$$x+y=2$$ is directrix of parabola

vertex is $$A(0,0)$$

latus rectum $$\Rightarrow 4a\Rightarrow 4\sqrt{2}.$$

eq. of parabola $$\Rightarrow $$

$$\left(\dfrac{x-y}{\sqrt{2}}\right)^{2}=4\sqrt{2}\left[\dfrac{x+y}{\sqrt{2}}\right]$$

$$(x-y)^{2}= 8 x+y$$

$$\Rightarrow x^{2}-2xy+y^{2}-8x- 8y=0$$


Class 11 Commerce Maths Extra Questions