Mathematics · Quantitative Aptitude

Conic Sections

239 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 parabola tracing of the parabola definitions related to parabola introduction to parabola

The parametric equation of a parabola is $x=t^{2}+1, y=2t+1$. The Cartesian equation of its directrix is 

  1. $x=0$
  2. $x+1=0$
  3. $y=0$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given x = t^2 + 1 and y = 2t + 1, we have t = (y - 1)/2. Substituting into x gives x = ((y - 1)/2)^2 + 1, or (y - 1)^2 = 4(x - 1). This is a parabola with vertex (1, 1), opening right, with 4a = 4, so a = 1. The directrix is x = h - a, which is x = 1 - 1 = 0.

Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola

$TP$ and $TQ$ are tangents to parabola $y^{2}=4x$ and normal at $P$ and $Q$ intersect at a point $R$ on the curve. The locus of the center of the circle circumscribing $\Delta TPQ$ is parabola whose

  1. Vertex is $\left(1,0\right)$.
  2. Foot of directrix is $\left(\dfrac{7}{8},0\right)$
  3. Length of latus-rectum is $\dfrac{1}{4}$.
  4. Focus is $\left(\dfrac{9}{8},0\right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the parabola y^2 = 4x, the locus of the circumcenter of triangle TPQ, where P and Q are points on the parabola and the normals at P and Q meet on the parabola, is a known property. The resulting locus is another parabola with vertex (1, 0).

Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola
Consider the conic $ x^{2}+4y-6x+k=0 $ & $\displaystyle L\Rightarrow y+1=0$ be its directrix
On the basis of above information answer the following question:

The focus of the parabola is

  1. $(-3, 3)$
  2. $(-3, -3)$
  3. $(3, -3)$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$x^2+4y-6x+k=0$
$\implies (x-3)^2=-4\left ( y-(\dfrac{k-9}{4}) \right )$
So, length of latus rectum $=4a=4$
$\therefore a=1$
Distance between vertex and directrix=a
$\therefore\left ( -1-(\dfrac{k-9}{4}) \right )=1$
$\therefore k=1$
So, co-ordinates of vertex are $\left(3,\dfrac{k-9}{4}\right)=(3,-2)$
Therefore, co-ordinates of focus are $(3,-2-a)=(3,-3)$
So, answer is option (C).
Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola

The vertex of the parabola $2((x-1)^2 + (y-2)^2) = (x + y + 3)^2$ is

  1. $\left (-\displaystyle \frac {1}{2}, -\frac {1}{2}\right )$
  2. $\left (-\displaystyle \frac {1}{2}, \frac {1}{2}\right )$
  3. $\left (\displaystyle \frac {1}{2}, \frac {1}{2}\right )$
  4. $\left (\displaystyle \frac {1}{2}, -\frac {1}{2}\right )$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given parabola may be written as,$\displaystyle \sqrt {(x-1)^2+(y-2)^2}=\frac {|x+y+3|}{\sqrt 2}$
$\Rightarrow$  focus is $S=(1,2)$, diretrix is, $x + y + 3 = 0 ....(1)$
We know axis of the parabola passes through focus and perpendicular to the directrix.
Thus equation of axis is,$x -y + 1 = 0 .....(2)$
solving (1) and (2) we get the foot of directrix $P(-2,-1)$
So the vertex of the parabola will be mid point of PS
$\Rightarrow V = \displaystyle \left (-\frac {1}{2}, \frac {1}{2}\right )$

Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola

If the vertex of the conic $y^{2} - 4y = 4x - 4a$ always lies between the straight lines $x + y = 3$ and $2x + 2y - 1 = 0$ then

  1. $2 < a < 4$
  2. $-\dfrac {1}{2} < a < 2$
  3. $0 < a < 2$
  4. $-\dfrac {1}{2} < a < \dfrac {3}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Vertex of $y^{2} - 4y = 4x - 4a$ is $(a - 1, 2)$
So, $(a - 1 + 2 - 3)(2a - 2 + 4 - 1) < 0$
$(a - 2)(2a + 1) < 0$
$-\dfrac {1}{2} < a < 2$

Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola

For the parabola $9x^{2} - 24xy + 16y^{2} - 20x - 15y - 60 = 0$ which of the following is/ are true.

  1. $focus = \left (-\dfrac {43}{25}, -\dfrac {129}{100}\right )$
  2. $focus = \left (\dfrac {43}{25}, \dfrac {129}{100}\right )$
  3. $directrix : 4x + 3y + \dfrac {53}{4} = 0$
  4. $directrix : 4x + 3y - \dfrac {53}{4} = 0$
Reveal answer Fill a bubble to check yourself
B,C Correct answer