Tag: conic sections

Quizzes (12) Questions (54)

Questions Related to conic sections

Multiple choice mathematics and statistics conic sections standard equation of ellipse introduction to ellipse standard equation of an ellipse

lf $ax^{2}+by^{2}+2gx+2fy+c=0$ represents an ellipse, then

  1. its major axis is parallel to $x-$axis
  2. its major axis is parallel to $y-$axis
  3. its axes (i.e. major axis and minor axis) are neither parallel to $x-$axis nor parallel to $y-$axis

  4. its axes are parallel to co-ordinate axes
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The general second-degree equation ax^2 + by^2 + 2gx + 2fy + c = 0 represents an ellipse with axes parallel to the coordinate axes because there is no xy term.

Multiple choice mathematics and statistics conic sections standard equation of ellipse introduction to ellipse standard equation of an ellipse

The abscissa of the focii of the ellipse $25(\mathrm{x}^{2}-6\mathrm{x}+9)+16\mathrm{y}^{2}=400$ is:

  1. $ (4,-ae), (4,ae)$
  2. $ (3,-ae), (3,ae)$
  3. $ (5,-ae), (5,ae)$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation of the ellipse can be written as $\dfrac{(x-3)^{2}}{4^{2}}+\dfrac{y^{2}}{5^{2}}=1$
$\therefore $ Minor axis is along the line $x-3=0$ and major axis is $y=0$ 
$(\because a^{2}=4^{2}< b^{2}=5^{2})$
$\therefore S\,$ and $ S^{'}$ are $(3,3),(3,-3)$
$\because 4^{2}=5^{2}(1-e^{2})\Rightarrow e=\dfrac{3}{5}\left ( \because ae=5\dfrac{3}{5}=3 \right )$ and $S\equiv(3,ae )$, $S^{'}\equiv(3,-ae).$
Ans: B

Multiple choice mathematics and statistics conic sections standard equation of ellipse introduction to ellipse standard equation of an ellipse

The eccentricity of the curve with equation ${ x }^{ 2 }+{ y }^{ 2 }-2x+3y+2=0$ is

  1. $0$
  2. $\sqrt { 2 }$
  3. $1/2$
  4. ${ 1 }/{ \sqrt { 2 } }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given ${x}^{2}+{y}^{2}-2x+3y+2=0$

$\Rightarrow \left({x}^{2}-2x\right)+\left({y}^{2}+3y\right)+2=0$
$\Rightarrow \left({x}^{2}-2x+1-1\right)+\left({y}^{2}+2\times 1\times \dfrac{3}{2}+\dfrac{9}{4}-\dfrac{9}{4}\right)+2=0$
$\Rightarrow {\left(x-1\right)}^{2}-1+{\left(y+\dfrac{3}{2}\right)}^{2}-\dfrac{9}{4}+2=0$
$\Rightarrow {\left(x-1\right)}^{2}+{\left(y+\dfrac{3}{2}\right)}^{2}=-2+1+\dfrac{9}{4}=\dfrac{5}{4}$
Divide both sides by ${\left(\sqrt{\dfrac{5}{4}}\right)}^{2}$ we get
$\dfrac{{\left(x-1\right)}^{2}}{{\left(\sqrt{\dfrac{5}{4}}\right)}^{2}}+\dfrac{{\left(y+\dfrac{3}{2}\right)}^{2}}{{\left(\sqrt{\dfrac{5}{4}}\right)}^{2}}$  is an ellipse where $a=\sqrt{\dfrac{5}{4}}$ and  $b=\sqrt{\dfrac{5}{4}}$
Eccentricity $e=\sqrt{1-\dfrac{{b}^{2}}{{a}^{2}}}=\sqrt{1-\dfrac{{\left(\sqrt{\dfrac{5}{4}}\right)}^{2}}{{\left(\sqrt{\dfrac{5}{4}}\right)}^{2}}}=\sqrt{1-1}=0$