Hyperbola: Definition, Equation, Formula and Sample Questions

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Hyperbola is the locus of all the points in a plane such that the difference in their distances from two fixed points in the plane is constant. Hyperbola is made up of two similar curves that resemble a parabola. Hyperbola has two fixed points which can be shown in the picture, are known as foci or focus. When we join the foci or focus using a line segment then its midpoint gives us centre. Hence, this line segment is known as the transverse axis.

Table of Content

Eccentricity of Hyperbola
Equation of Tangents
Sample Questions

Eccentricity of Hyperbola

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The distance ratio from the centre to a vertex and from centre to a focus is called the eccentricity of Hyperbola.

Eccentricity (e) =c/a;

Eccentricity is always positive for hyperbola, since c>=a.

Hyperbola

Standard Equation for Hyperbola

Let us now derive the standard equation of hyperbola. For this, consider a hyperbola with center O at(0,0) and its foci lie on any one of the x or y axis.

Both the foci’s lie at a distance of “c” on the x-axis and the vertices are at a distance “a” from (0,0) origin. Let us consider a point Z on the Hyperbola so that it satisfies the definition ZF1+zF2 is constant 2a.

2a= zF1+zF2

According to the distance formulae,

√(x+c)^2+y^2 -√(x-c)^2+y^2 =2a ……..eqn(1)

√(x+c)^2+y^2=2a+√(x-c)^2+y^2

Squaring both sides,

X^2/a^2-y^2/c^2-a^2=1

Since we know that b^2=c^2-a^2 and 0<a<c

y^2=b^2(x^2/a^2-1)

Substituting y^2 in eqn(1):

X^2/a^2-y^2/b^2=1

Latus rectum of Hyperbola

The line segments perpendicular to the transverse axis through any of the foci such that their endpoints lie on the hyperbola are defined as the latus rectum of a hyperbola.

The length of the latus rectum is 2b²/a. The ends of the latus rectum of a hyperbola are (ae,+-b^2/a^2).

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Conic Sections Detailed Video Explanation:

Equation of Tangents

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The equation of tangents to the hyperbola at the point(x1,y1) is xx1/a^2-yy1/b^2=1

In general, two tangents can be drawn to the hyperbola from an external point (x1,y1) to the hyperbola and they are given by the equations (y-y1) =m(x-x1) and (y-y2)=m(x-x2),where m1 and m2 are the roots of the equation (x1^2-a^2)m^2-2x1y1m+y1^2+b^2=0

In parametric coordinates the equation of tangent is given by xsecθ/a - ytan(θ/b)=1

Equation of Tangents

Equation of Normal of Hyperbola

The equation of the normal to Hyperbola at the point (x1,y1) is a^2x/x1+b^2y/y1=a^2-b^2=a^2e^2.

In parametric coordinates , the equation becomes ax/secΘ+by/tanΘ=a^2+b^2=a^2e^2.

Equation of Normal of Hyperbola

Sample Questions

Ques: Differentiate between hyperbola and Parabola? (2 marks)

Ans. When we observe the meaning of hyperbola,it seems precisely like two parabola’s opposite to each other. Hyperbola and parabola are very much different from each other. In hyperbola, we have to calculate how distant a set of points is from two fixed points where in parabola we have to calculate that the set of points are from the same distance from the directrix. One difference is that all parabola are of the same shape but hyperbola are of different shapes. Parabola’s two arms are parallel but in hyperbola they are not parabola.

Ques: How to calculate the length of the latus rectum in hyperbola? (1 mark)

Ans. Latus rectum is a chord parallel to the directrix and passing through focus. One can observe that hyperbola has two latus rectums. b^2/a is the length of the semi latus rectum and 2(b^2/a) gives the full length of the latus rectum.

Ques: The locus of a point p(a,b) moving under the condition that the line y=ax+b is a tangent to the hyperbola x^2/a^2-y^2/b^2=1 is. (1 mark)

Ans. Tangent to the hyperbola x^2/a^2-y^2/b^2=1 is y=mx+_√(a^2m^2-b^2).

Given that y=ax+b is the tangent of the hyperbola. m=a and a^2m^2-b^2=b^2. Locus is a^2x^2-y^2=b^2 which is parabola.

Ques: If a hyperbola passes through the point P(10,16) and it has vertices at(+-6,0) then the equation of the normal at p is : (2 marks)

Ans. Vertex of hyperbola is =6

We know the eqn of hyperbola is x^2/a^2-y^2/b^2=1.

X^/36-y^2/b^2=1

Point p(10,16) lies on parabola,so

100/36 -256/b^2=1

So, 64/36=256/b^2

Therefore, b^2=144. So, eqn. of hyperbola becomes x^2/36-y^2/144 =1 and eqn of normal : 2x+5y=100.

Ques: If e1 and e2 are the eccentricities of the ellipse x^2/18+y^2/4=1 and the hyperbola x^2/9-y^/4 =1 respectively and (e1,e2) is a point on the eclipse 15x^2+3y^2=k then k is equal to? (1 mark)

Ans. e1 and e2 are the eccentricities of the ellipse x^2/18 +y^/4=1 and the hyperbola x^/9-y^/4 =1 respectively.

e1=√7/3 and e2=√13/3 as(e1,e2) lies on the ellipse 15x^2 +3y^2=k because, 15e1^2+3e2^2=k so therefore k=16.

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