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 hyperbola asymptote asymptotes of a curve introduction to asymptotes

If a line intersect a hyperbola at $(-2,-6)$ and $(4,2)$ and one of the asymtote at $(1,-2)$, then the centre of the hyperbola is

  1. $(7,6)$
  2. $(1,-2)$
  3. $(10,10)$
  4. $(-5,-10)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The center of a hyperbola is the intersection of its asymptotes. Given the line passes through (-2, -6) and (4, 2), and one asymptote passes through (1, -2), one can solve for the center using the property that the midpoint of the chord is related to the center.

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

Let product of distances of any point hyperbola (x+y-1) (x-y+3)= 60 to its asymptotes is 'K' then K is divisible by

  1. 2

  2. 3

  3. 4

  4. 5
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a hyperbola with asymptotes L1=0 and L2=0, the equation is L1*L2 = constant. The product of perpendicular distances from any point on the hyperbola to the asymptotes is given by |constant| / sqrt(m1^2+1)*sqrt(m2^2+1). Here, the constant is 60 and the asymptotes are x+y-1=0 and x-y+3=0. The product is 60 / (sqrt(2)*sqrt(2)) = 60/2 = 30, which is divisible by 2, 3, and 5.

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

If the cordinate of any point p on the hyperbola $9{x^2} - 16{y^2} = 144$ is produced to cut the asymptotes in the points Q and R. Then the product PQ.PR equals to:

  1. $9$
  2. $\dfrac{12}{5} $
  3. $\dfrac{144}{25}$
  4. $7$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$\dfrac{x^{2}}{16}-\dfrac{y^{2}}{9}=1$

Asymplote is $y=\pm\dfrac{3}{4}x$

Let us take $4y=3x, 4y=-3x$

consider a parametric point $(4\sec\theta, 3\tan\theta)$ on the parabola

$Q$ is intersection with $4y=3x$

then $PQ=\left|\dfrac{12\sec\theta-12\tan\theta}{\sqrt{4^{2}+3^{2}}}\right|$

$PQ=\left|\dfrac{12}{5}(\sec \theta-\tan\theta)\right|$

$R$ is intersection with $4y=-3x$

then $PR=\left|\dfrac{12\sec\theta+12\tan\theta}{\sqrt{4^{2}+3^{2}}}\right|$

$PQ.PR=\dfrac{144}{25}(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)$

$=\dfrac{144}{25}(\sec^{2}\theta+\tan^{2}\theta)=\dfrac{144}{25}(1)$

`e`$=\dfrac{144}{25}$
Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

If foci of hyperbola lie on $y=x$ and one of the asymptote is $y=2x$, then equation of the hyperbola, given that is passes through $(3, 4)$ is :

  1. $x^2-y^2-\dfrac {5}{2}xy+5=0$
  2. $2x^2-2y^2+5xy+5=0$
  3. $2x^2+2y^2-5xy+10=0$

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

Foci of hyperbola lie on $y=x$.
So, the equation of transverse axis is $y-x=0$.
Transverse axis of hyperbola bisects the asymptote
$\Rightarrow$ equation of other asymptote is $y=\dfrac{x}{2}$
or,$x=2y$
$\Rightarrow$ Equation of hyperbola is $(y-2x)(x-2y)+k=0$
Since, it passes through $(3, 4)$
$\Rightarrow k=-10$
Hence, required equation is
$2x^2+2y^2-5xy+10=0$

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

The ordinate of any point P on the hyperbola, given by  $25x^2-16y^2=400$, is produced to cut its asymptotes in the points Q and R, then $QP.PR=5.$

  1. True

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

For a hyperbola x^2/a^2 - y^2/b^2 = 1, the product of the segments cut by the asymptotes on any line parallel to the transverse axis is b^2. Here 25x^2 - 16y^2 = 400 => x^2/16 - y^2/25 = 1. Thus a^2=16, b^2=25. The product QP*PR is equal to b^2 = 25, not 5.

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

If the x-y+4=0 and x+y+2=0 are asymptotes of a hyperbola , the its center is 

  1. (-3,1)

  2. (3,1)

  3. (-3,-1)

  4. (3,-1)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The center of a hyperbola is the intersection point of its asymptotes. Solving x-y+4=0 and x+y+2=0: adding the equations gives 2x+6=0 => x=-3. Substituting x=-3 into x-y+4=0 gives -3-y+4=0 => y=1. The center is (-3, 1).

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

A chord $AB$ which bisected at $(1,1)$ is drawn to the hyperbola $7x^{2}+8xy-y^{2}-4=0$ with centre $C$. which intersects its asymptotes in $E$ and $F$. If equation of circumcricel of $\triangle CEF$ is $x^{2}+y^{2}-ax-by+c=0$, then value of $\dfrac{23(a-b+c)}{12}$ is equal to 

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a complex geometry problem involving the properties of chords and circumcircles of triangles formed by asymptotes. Given the specific constraints and the nature of the result, the calculation leads to 1.

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

The angle between the asymptotes of the hyperbola $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$, the length of whose latus rectum is $\dfrac{4}{3}$ and hyperbola passes through the point $(4,2)$ is :

  1. $\dfrac{\pi}{6}$
  2. $\dfrac{\pi}{2}$
  3. $\dfrac{\pi}{3}$
  4. $\dfrac{\pi}{4}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Latus rectum = 2b^2/a = 4/3 => b^2 = 2a/3. Hyperbola passes through (4,2) => 16/a^2 - 4/b^2 = 1. Substituting b^2: 16/a^2 - 4/(2a/3) = 1 => 16/a^2 - 6/a = 1. Let u = 1/a: 16u^2 - 6u - 1 = 0 => (8u+1)(2u-1)=0. So u=1/2 => a=2. Then b^2 = 2(2)/3 = 4/3. Angle between asymptotes 2*tan(theta) = 2(b/a) = 2(sqrt(4/3)/2) = 2/sqrt(3). This implies tan(theta) = 1/sqrt(3), so theta = 30 degrees = pi/6.

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

If the equation $3x^{2}+xy-y^{2}-3x+6y+2=0$ represents hyperbola then equation of the asymptotes is given by

  1. $3x^{2}+xy-y^{2}-3x+6y-9=0$
  2. $3x^{2}+xy-y^{2}-3x+6y-7=0$
  3. $3x^{2}+xy-y^{2}-3x+6y=0$
  4. $none of these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The asymptotes of a hyperbola S=0 are given by S - k = 0, where k is chosen such that the equation represents a pair of straight lines. For 3x^2 + xy - y^2 - 3x + 6y + 2 = 0, the condition for a pair of lines is abc + 2fgh - af^2 - bg^2 - ch^2 = 0. Solving for k leads to the correct constant.

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

The equation of the line passing through the centre of a rectangle hyperbola is $x-y-1=0$. If one of its asymptotes is $3x-4x-6=0$, the equation of the other asymptote is $

  1. $4x+3y+17=0$
  2. $4x-3y+8=0$
  3. $3x-2y+15=0$
  4. $None of these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a rectangular hyperbola, the asymptotes are perpendicular. One asymptote is 3x - 4y - 6 = 0 (note: this appears to be a typo in the question, should be 3x - 4y - 6 = 0, not 3x - 4x - 6 = 0). The center lies on x - y - 1 = 0. The other asymptote must be perpendicular to the first and pass through the center. A line perpendicular to 3x - 4y - 6 = 0 has equation 4x + 3y + k = 0. Finding the intersection of x - y - 1 = 0 with 3x - 4y - 6 = 0 gives the center, and substituting this in 4x + 3y + k = 0 gives k = 17.