Mathematics · Quantitative Aptitude

Conic Sections

245 Questions

Conic sections deal with the curves obtained by the intersection of a cone with a plane, primarily focusing on parabolas, ellipses, and hyperbolas. Questions require finding vertices, directrices, and asymptotes based on given equations. This is an advanced geometry topic for rigorous competitive exams.

Parabola equationsHyperbola propertiesEllipse conceptsTangents and normalsDirectrix and focus

Conic Sections Questions

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

An ellipse $\cfrac { { x }^{ z } }{ 4 } +\cfrac { { y }^{ z } }{ 3 } =1$ confocal with hyperbola $\cfrac { { x }^{ 2 } }{ \cos ^{ 2 }{ \theta  }  } -\cfrac { { y }^{ 2 } }{ \sin ^{ 2 }{ \theta  }  } =1$ then the set of value of $'0'$

  1. $R$
  2. $R-\left\{ n\pi ,n\epsilon z \right\} $
  3. $R-\left\{ \left( 2n+1 \right) \cfrac { \pi }{ 2 } ,n\epsilon z \right\} $
  4. $R-\left\{ \cfrac { n\pi }{ 2 } ,n\epsilon z \right\} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Focus of ellipse$=ae=a\sqrt { 1-\cfrac { { b }^{ 2 } }{ { a }^{ 2 } }  } $
$=\sqrt { { a }^{ 2 }-{ b }^{ 2 } } $
$=\sqrt { 1 } $
$=1$
Focus of hyperbola$=a\sqrt { 1+\cfrac { { b }^{ 2 } }{ { a }^{ 2 } }  } $
$=\sqrt { { a }^{ 2 }+{ b }^{ 2 } } $
$=\sqrt { \sin ^{ 2 }{ \theta  } +\cos ^{ 2 }{ \theta  }  } $
$=\sqrt { 1 } $
$=1$
$\therefore $The ellipse and hyperbola will be confocal for $\theta \epsilon R$.
Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

If equation $(5x-1)^{2}+(5y-2)^{2}=(\lambda^{2}-2\lambda+1)(3x+4y-1)^{2}$ represents an ellipse, then $\lambda \in$

  1. $(0, 1)$
  2. $(0, 2)$
  3. $(1, 2)$
  4. $(0, 1)\cup (1, 2)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation represents an ellipse if the eccentricity e < 1. This condition relates to the coefficients of the quadratic form.

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

The locus of the mid points of the portion of the tangents to the ellipse intercepted between the axes

  1. $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=4$
  2. $\dfrac{a^{2}}{x^{2}}+\frac{b^{2}}{y^{2}}=4$
  3. $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=4$

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

Let a tangent to the ellipse x^2/a^2 + y^2/b^2 = 1 be x/a cos(theta) + y/b sin(theta) = 1. The intercepts on the axes are a/cos(theta) and b/sin(theta). If (h, k) is the midpoint of this intercepted portion, then h = a / (2 cos(theta)) and k = b / (2 sin(theta)). Eliminating theta gives a^2/h^2 + b^2/k^2 = 4.

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

The length of the latusrectum of the parabola $169\left{ { \left( x-1 \right)  }^{ 2 }+{ \left( y-3 \right)  }^{ 2 } \right} ={ \left( 5x-12y+17 \right)  }^{ 2 }$

  1. $\cfrac { 14 }{ 13 } $
  2. $\cfrac { 28 }{ 13 } $
  3. $\cfrac { 12 }{ 13 } $

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

Here ${ \left( x-1 \right)  }^{ 2 }+{ \left( y-3 \right)  }^{ 2 }={ \left{ \cfrac { 5x-12y+17 }{ \sqrt { { 5 }^{ 2 }+{ \left( -12 \right)  }^{ 2 } }  }  \right}  }^{ 2 }$


$\therefore$ The focus is $(1,3)$ and the directrix is $5x-12y+17=0$

The distance of the focus from the directrix

$=\left| \cfrac { 5\times 1-12\times 3+17 }{ \sqrt { { 5 }^{ 2 }+{ \left( -12 \right)  }^{ 2 } }  }  \right| =\cfrac { 14 }{ 13 } $

$\therefore$ Length of latusrectum $=2\times \cfrac { 14 }{ 13 } =\cfrac { 28 }{ 13 } $

Multiple choice maths constructions mid-point formula midpoints division of a line segment

If Q is a variable point on $x^2=4y$ and O is the origin, the locus of mid point OQ is equation of 

  1. an ellipse

  2. a parabola

  3. hyperbola

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

Let Q be (t^2, 2t) on the parabola x^2 = 4y. The midpoint OQ is (h, k) = (t^2 / 2, t / 2). Eliminating the parameter t gives t = 2k, so h = (2k)^2 / 2 = 2k^2, which rearranges to y^2 = (1/2)x, representing another parabola.

Multiple choice maths constructions mid-point formula midpoints division of a line segment

The locus of the mid-point of that chord of parabola which subtends right angle on the vertex will be

  1. $y ^ { 2 } - 2 a x + 8 a ^ { 2 } = 0$
  2. $y ^ { 2 } = a ( x - 4 a )$
  3. $y ^ { 2 } = 4 a ( x - 4 a )$
  4. $y ^ { 2 } + 3 a x + 4 a ^ { 2 } = 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a parabola y^2 = 4ax, the chord subtending a right angle at the vertex has the equation y = mx + 2am. The midpoint (h, k) of this chord satisfies k = mh + 2am and the property that the chord is y = (2a/k)x - 4a^2/k. Substituting and simplifying leads to the locus y^2 = 2a(x - 4a), which is y^2 - 2ax + 8a^2 = 0.

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

If line $y+3x=c$ is normal of the ellipse ${ x }^{ 2 }+3{ y }^{ 2 }=3$ then equation of normal is-

  1. $y-3x\pm \sqrt { 3 } =0$
  2. $y+3x\pm \sqrt { 3 } =0$
  3. $y+3x\pm 3 =0$
  4. $y+3x\pm 1 =0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If $y=mx+c$ is normal to ellipse
${ c }^{ 2 }={ m }^{ 2 }\cfrac { { \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 }{ m }^{ 2 } } $
${ x }^{ 2 }+3{ y }^{ 2 }=3$
$\cfrac { { x }^{ 2 } }{ 3 } +\cfrac { { y }^{ 2 } }{ 12 } =1$
${ a }^{ 2 }=3,b=1$
$y+3x=c$
$m=-3$
${ c }^{ 2 }={ (-3) }^{ 2 }\cfrac { { \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 }{ m }^{ 2 } } =9\times \cfrac { { (3-1) }^{ 2 } }{ 3+9 } $
$=\cfrac { 9\times 4 }{ 12 } =3$

Equation of normal
$y+3x\pm \sqrt { 3 } =0$

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

If the tangent drawn at a point $\left( { t }^{ 2 },2t \right) $ on the parabola ${ y }^{ 2 }=4x$ is same as normal drawn at $\left( \sqrt { 5 } \cos { \alpha  } ,2\sin { \alpha  }  \right) $ on the ellipse $\displaystyle \frac { { x }^{ 2 } }{ 5 } +\frac { { y }^{ 2 } }{ 4 } =1$, then which of following is true.

  1. $\displaystyle t=\pm \frac { 1 }{ \sqrt { 5 } } $
  2. $\alpha =-\tan ^{ -1 }{ 2 } $
  3. $\alpha =\tan ^{ -1 }{ 2 } $

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

Equation of tangent to ${ y }^{ 2 }=4x$ at $\left( { t }^{ 2 },2t \right) $ is $x=ty-{ t }^{ 2 }$   ....(1)


Equation of normal to ellipse $\displaystyle \frac { { x }^{ 2 } }{ 5 } +\frac { { y }^{ 2 } }{ 4 } =1$ at $\left( \sqrt { 5 } \cos { \alpha  } ,2\sin { \alpha  }  \right) $ is $\sqrt { 5 } \sec { \alpha x-2y\csc { \alpha  } =1 } $    ....(2)

Given (1) $=$ (2)

$\displaystyle \Rightarrow \sqrt { 5 } \sec { \alpha  } =\frac { 2\csc { \alpha  }  }{ t } =-\frac { 1 }{ { t }^{ 2 } } \Rightarrow \cos { \alpha  } =-\sqrt { 5 } { t }^{ 2 }$ and $\sin { \alpha  } =-2t$

$\displaystyle \Rightarrow \cos ^{ 2 }{ \alpha  } +\sin ^{ 2 }{ \alpha  } =5{ t }^{ 4 }+4{ t }^{ 2 }=1\Rightarrow { t }^{ 2 }=\frac { 1 }{ 5 } $   

$\therefore$ (A) is true
and $\displaystyle \frac { \sin { \alpha  }  }{ \cos { \alpha  }  } =-\frac { 2t }{ -\sqrt { 5 } { t }^{ 2 } } =\frac { 2 }{ \sqrt { 5 }  } \times \frac { 1 }{ t } =\frac { 2 }{ \sqrt { 5 }  } \times \left( \pm 5 \right) $

$\therefore \tan { \alpha  } =\pm 2$

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

lf the tangent drawn at a point $(t^{2},2t)$ on the parabola $y^{2}=4x$ is same as normal drawn at $(\sqrt{5}\cos\alpha, 2\sin\alpha)$ on the ellipse $\displaystyle \frac{x^{2}}{5}+\frac{y^{2}}{4}=1$, then which of following is not true?  

  1. $t=\displaystyle \pm\frac{1}{\sqrt{5}}$
  2. $\alpha=-\tan^{-1}2$
  3. $\alpha=\tan^{-1}2$
  4. $\alpha=\tan^{-1}4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The tangent to y^2=4x at (t^2, 2t) is ty = x + t^2. The normal to x^2/5 + y^2/4 = 1 at (sqrt(5)cos(alpha), 2sin(alpha)) is sqrt(5)xsec(alpha) - 2ycosec(alpha) = 1. Comparing coefficients leads to the condition for the lines to be identical, which excludes option D.

Multiple choice

The Ajanta Caves, located in Maharashtra, India, are known for their Buddhist rock-cut cave temples and paintings. What is the mathematical shape of the caves' arched entrances?

  1. Circle

  2. Ellipse

  3. Parabola

  4. Hyperbola
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The arched entrances of the Ajanta Caves are elliptical in shape, which is a two-dimensional shape defined by the intersection of a plane and a cone.

Multiple choice

Which conic section is represented by the equation (y^2 = 4px)?

  1. Circle

  2. Ellipse

  3. Parabola

  4. Hyperbola
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation (y^2 = 4px) represents a parabola with vertex at the origin and axis of symmetry along the (x)-axis.

Multiple choice

What is the standard form of the equation of a parabola?

  1. \(x^2 + y^2 = r^2\)
  2. \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)
  3. \(y^2 = 4px\)
  4. \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The standard form of the equation of a parabola is (y^2 = 4px), where (p) is the distance from the vertex to the focus.