Tag: conic sections

$9x^2+5(y^2-6y)=0$
$9x^2+5(y^2-6y+9)=45$
$9x^2+5(y-3)^2=45$
$\dfrac { x^2 }{ 5 }+\dfrac { (y-3)^2 }{ 9 }=1$  
$a^2 < b^2$
$a^2=b^2(1-e^2)$
$5=9(1-e^2)$
$\dfrac { 5 }{ 9 }=1-e^2$
$e^2=\dfrac{4}{9}$ 
$e=\dfrac{2}{3}$

The vertices are (2, -2) and (2, 4), so the center is (2, 1) and the major axis is vertical with length 2a = 6, so a = 3. Given e = a/3 = 1, this is a degenerate case or the eccentricity is defined differently. Assuming standard ellipse properties, the vertical orientation and center (2, 1) match option B.

We know that,

The equation of ellipse whose centre $(h, k)$.

$\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1$

Given centre of ellipse $\left( h,k \right)=\left( 1,2 \right)$

Then, equation of ellipse $\dfrac{{{\left( x-1 \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-2 \right)}^{2}}}{{{b}^{2}}}=1$          ……. (1)

But, the ellipse passes through given point $\left( x,y \right)=\left( 4,6 \right)$

By equation (1), we get

$ \dfrac{{{\left( 4-1 \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( 6-2 \right)}^{2}}}{{{b}^{2}}}=1 $

$ \dfrac{{{3}^{2}}}{{{a}^{2}}}+\dfrac{{{4}^{2}}}{{{b}^{2}}}=1 $

$ \Rightarrow 9{{b}^{2}}+16{{a}^{2}}={{a}^{2}}{{b}^{2}} $

$\Rightarrow 16{{a}^{2}}+9{{b}^{2}}={{a}^{2}}{{b}^{2}}$        …… (2)

Now, distance between focus and centre is $c=\sqrt{{{a}^{2}}-{{b}^{2}}}$

So,

$ c=\sqrt{{{\left( 1-6 \right)}^{2}}+{{\left( 2-2 \right)}^{2}}} $

$ c=\sqrt{{{5}^{2}}} $

$ c=5 $

$\sqrt{{{a}^{2}}-{{b}^{2}}}=5$

On squaring both sides, we get,

${{a}^{2}}-{{b}^{2}}=25$          …… (3)

By equation (2) and (3), we get

$ 9{{b}^{2}}+400+16{{b}^{2}}=25{{b}^{2}}+{{b}^{4}} $

$ 25{{b}^{2}}+400=25{{b}^{2}}+{{b}^{4}} $

$ 400={{b}^{4}} $

$ {{b}^{4}}-400=0 $

$ \left( {{b}^{2}}-20 \right)\left( {{b}^{2}}+20 \right)=0 $

${{b}^{2}}-20=0\,$            and        ${{b}^{2}}+20=0\,$

${{b}^{2}}=20$                and        ${{b}^{2}}=-20$      (Rejected)

Put the value of ${{b}^{2}}$ in equation (3) and we get,

${{a}^{2}}-{{b}^{2}}=25$

$ {{a}^{2}}-20=25 $

$ {{a}^{2}}=45 $

Now, put the value of ${{a}^{2}}$ and ${{b}^{2}}$ in equation (1), we get,

$\dfrac{{{\left( x-1 \right)}^{2}}}{45}+\dfrac{{{\left( y-2 \right)}^{2}}}{20}=1$

$3x^2+2y^2+6x-8y+5=0$
$\Rightarrow \dfrac {(x+1)^2}{2}+\dfrac {(y-2)^2}{3}=1$
Therefore, centre is $(-1, 2)$ and ellipse is vertical
$(\because b > a)$
$a^2=2, b^2=3$
Now $2=3(1-e^2)$
$\Rightarrow e=\dfrac {1}{\sqrt 3}$
Foci are $(-1, 2\pm be)$ and $(-1, 2\pm 1)$.
Hence, the foci are $(-1, 3)$ and $(-1, 1)$
The equations of the directrices are $y=2\pm \dfrac {b}{e}\Rightarrow y=5$ and $y=-1$

Given ellipse may be written as,
$\displaystyle9x^{2}+4\left ( y^{2}-\frac{15}{2}y+\frac{225}{16} \right

)=\frac{225}{4}$
or $\displaystyle \frac{x^{2}}{225/36}+\frac{\left (

y-15/4 \right )^{2}}{225/16}=1$
$\displaystyle \therefore

b^{2}=\frac{225}{16}, a^{2}=\frac{225}{36}$
$\displaystyle \therefore

e^{2}=1-\frac{a^{2}}{b^{2}}=1-\frac{16}{36}=1-\frac{4}{9}=\frac{5}{9}$
$\displaystyle

\therefore e=\frac{1}{3}\sqrt{5}.$

$\displaystyle 4x^{2}+y^{2}-8x+2y+1=0$
$\displaystyle 4(x^2-2x)+(y^2+2y)=-1$
$\displaystyle 4(x^2-2x+1)+(y^2+2y+1)=-1+5=4$
$\displaystyle 4(x-1)^2+(y+1)^2=4$
$\displaystyle \frac{\left ( x-1 \right )^{2}}{1}+\frac{\left (

y+1 \right )^{2}}{4}=1$
$\displaystyle \Rightarrow a^2 =1, b^2 = 4$ Clearly here $a^2< b^2$ so the axis of the ellipse is parallel to y-axis
Now,  $\displaystyle e=\sqrt{1-\frac{a^2}{b^2}}=\frac{\sqrt{3}}{2} \therefore $ Foci $\displaystyle \left

( 1, -1\pm \sqrt{3} \right ),$ Directrices  $\displaystyle y=-1\pm

\frac{4}{\sqrt{3}}.$

$4x^2+9y^2+8x+36y+4=0$
$\Rightarrow 4(x^2+2x)+9(y^2+4y)=-4$
$\Rightarrow 4(x^2+2x+1)+9(y^2+4y+4)=-4+4+36$
$\Rightarrow 4(x+1)^2+9(y+2)^2=36$
$\Rightarrow \dfrac{(x+1)^2}{9}+\dfrac{(y+2)^2}{4}=1$
$\therefore a^2=9,b^2=4$
Hence length of latus rectum is $=\dfrac{2b^2}{a}=\dfrac{8}{3}$

$12x^{ 2 }+4y^{ 2 }+24x-16y+25=0$