Tag: conic sections

Quizzes (12) Questions (54)

Questions Related to conic sections

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

The exhaustive interval of $\lambda$ for which the equation $\dfrac{x^2}{(\lambda^2-2\lambda-3)}+\dfrac{y^2}{\lambda^2+2\lambda-8}=1$ represents a hyperbola is 

  1. $ \lambda \varepsilon (- \infty, -4) \cup (3, \infty)$
  2. $ \lambda \varepsilon (-4, -1) \cup (2, 3)$
  3. $ \lambda \varepsilon (- \infty, -1) \cup (2, \infty)$
  4. $ \lambda \varepsilon (-4,-1)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

If area of quadrilateral formed by tangents drawn at ends of latus rectum of hyperbola $\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } -\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ is equal to square of distance between centre and one focus of hyperbola,then ${ e }^{ 3 }$ is (e is eccentricity of hyperbola)

  1. $2\sqrt { 2 } $

  2. 2

  3. 3

  4. 8
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The area of the quadrilateral formed by tangents at the ends of the latus rectum is 2*b^2. Setting this equal to (ae)^2 and using b^2 = a^2(e^2 - 1) leads to e^3 = 2*sqrt(2).

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

Eccentricity of a hyperbola is always less than 1.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Standard equation of the hyperbola is $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}$

The eccentricity of the hyperbola is given by
$e=\sqrt { 1+\dfrac { { b }^{ 2 } }{ { a }^{ 2 } }  }$ which is always greater than $1$.
Thus, the given statement is false.
Hence, option B is correct.

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

Equation of the latus rectum of the hyperbola $(10x - 5)^{2} + (10y - 2)^{2} = 9(3x + 4y - 7)^{2}$ is

  1. $y - 1/5 =-3/4(x - 1/2)$
  2. $x - 1/5 =-3/4(y - 1/2)$
  3. $y + 1/5 =-3/4(x + 1/2)$
  4. $x + 1/5 =-3/4(y + 1/2)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation is of the form (distance from focus)^2 = e^2 * (distance from directrix)^2. By identifying the focus and directrix, one can determine the latus rectum equation.

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

The equation $\frac{x^2}{1-k}-\frac{y^2}{1+k}=1$, $k<1$ represents 

  1. $circle$
  2. $ellipse$
  3. $hyperbola$
  4. $none$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Equating the above equation with the second-degree equation
$A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$ with $\dfrac{{x}^{2}}{1-k}-\dfrac{{y}^{2}}{1+k}=1$
we get $A=\dfrac{1}{1-k}, B=0, C=\dfrac{1}{1+k},D=0,E=0$ and $F=-1$
$(i)$For the second degree equation to represent a circle , the coefficients must satisfy the discriminant condition ${B}^{2}-4AC=0$ and also $A=C$
$\Rightarrow -4\times \dfrac{1}{1-k}\times \dfrac{1}{1+k}=0$
$\Rightarrow \dfrac{1}{1-{k}^{2}}=0$
This case does not exist
$(ii)$For the second degree equation to represent a ellipse , the coefficients must satisfy the discriminant condition ${B}^{2}-4AC<0$ and also $A\neq C$
$\Rightarrow -4\times \dfrac{1}{1-k}\times \dfrac{1}{1+k}<0$
$\Rightarrow \dfrac{1}{1-{k}^{2}}>0$
$\Rightarrow 1-{k}^{2}<0$
$\Rightarrow -{k}^{2}<-1$
$\Rightarrow {k}^{2}>1$ does not exist since it is given that $k<1$
$(iii)$For the second degree equation to represent a hyperbola, the coefficients must satisfy the discriminant condition ${B}^{2}-4AC>0$ and also $A\neq C$
$\Rightarrow -4\times \dfrac{1}{1-k}\times \dfrac{1}{1+k}>0$
$\Rightarrow \dfrac{1}{1-{k}^{2}}<0$
$\Rightarrow 1-{k}^{2}>0$
$\Rightarrow -{k}^{2}>-1$
$\Rightarrow {k}^{2}<1$ 
$\therefore k<1$
Hence the above equation represents a hyperbola.
Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

The equation $\displaystyle\frac{x^2}{10-\lambda}+\frac{y^2}{6-\lambda}=1$ represents

  1. a hyperbola if $\lambda < 6$
  2. an ellipse if $\lambda>6$
  3. a hyperbola if $6 < \lambda < 10$
  4. an ellipse if $0 < \lambda < 6$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

The general equation of an ellipse is $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ and that of a hyperbola is $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$
Using this, we get that the above equation is an ellipse if $10 - \lambda > 0$ and $6 - \lambda > 0$. The combined solution gives $\lambda < 6.$
For a hyperbola, the coefficient of $x^2$ and $y^2$ must be of opposite sign. Hence, 
 $10 - \lambda > 0$ and $6 - \lambda < 0$ which gives $6 < \lambda < 10$

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

The point to which the axes are to be translated to eliminate $x$ and $y$ terms in the equation $3x^{2}-4xy-2y^{2}-3x-2y-1=0$ is 

  1. $\left(\dfrac{5}{2},3\right)$
  2. $(-4,\dfrac{3}{2})$
  3. $ (-2,3)$
  4. $ (2,3)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given equation is $3x^{2}-4xy-2y^{2}-3x-2y-1=0$

Let $\left({x} _{1},{y} _{1}\right)$ be a point to which the origin is shifted by translation

Let $\left(X,Y\right)$ be the new coordinates of the point $\left(x,y\right)$

$\therefore\,$ the equations of the transformation are $x=X+{x} _{1},\,y=Y+{y} _{1}$

Now the transformed equation is 
$3{\left(X+{x} _{1}\right)}^{2}-4\left(X+{x} _{1}\right)\left(Y+{y} _{1}\right)-2{\left(Y+{y} _{1}\right)}^{2}-3\left(X+{x} _{1}\right)-2\left(Y+{y} _{1}\right)-1=0$

$\Rightarrow\,3\left({X}^{2}+2X{x} _{1}+{{x} _{1}}^{2}\right)-4\left(XY+X{y} _{1}+Y{x} _{1}+{x} _{1}{y} _{1}\right)-2\left({Y}^{2}+{{y} _{1}}^{2}+2Y{Y} _{1}\right)-3X-{x} _{1}-2Y-2{y} _{1}-1=0$

$\Rightarrow\,3{X}^{2}+3{{x} _{1}}^{2}+6X{x} _{1}-4X{y} _{1}-4{x} _{1}Y-4{x} _{1}{y} _{1}-2{Y}^{2}-2{{y} _{1}}^{2}+4Y{y} _{1}-3X-3{x} _{1}-2Y-2{y} _{1}-1=0$

$\Rightarrow\,\left(3{X}^{2}-4XY-2{Y}^{2}\right)+\left(3{{x} _{1}}^{2}-2{{y} _{1}}^{2}-4{x} _{1}{y} _{1}-3{x} _{1}-2{y} _{1}-1\right)+2X\left(3{x} _{1}-2{y} _{1}-\dfrac{3}{2}\right)+2Y\left(-2{x} _{1}+2{y} _{1}-1\right)=0$

Solving the first degree terms,we have
$3{x} _{1}-2{y} _{1}=\dfrac{3}{2}$

$-2{x} _{1}+2{y} _{1}=1$

Adding the above equations, we get
$3{x} _{1}-2{y} _{1}-2{x} _{1}+2{y} _{1}=\dfrac{3}{2}+1$

$\Rightarrow\,{x} _{1}=\dfrac{5}{2}$

From equation ,$-2{x} _{1}+2{y} _{1}=1$

$\Rightarrow\,2{y} _{1}=1+2{x} _{1}=1+2\times\dfrac{5}{2}=1+5=6$

$\Rightarrow\,{y} _{1}=\dfrac{6}{2}=3$

$\therefore\,\left({x} _{1},{y} _{1}\right)=\left(\dfrac{5}{2},3\right)$


Hence the point is $\left(\dfrac{5}{2},3\right)$