Find the equation of the hyperbola, whose foci are ± 5 , 0 and length of transverse axis is 8 .

Given foci = ± 5 , 0 and length of transverse axis is 8 . Since foci lie on the x - axis then the equation of the hyperbola is of the form x 2 a 2 - y 2 b 2 = 1 . W know that length of transverse axis is 2 a = 8 ∴ a = 4 We have foci = ± a e , 0 = ± 5 , 0 ∴ a e = 5 ⇒ e = 5 4 We know that e = a 2 + b 2 a ⇒ 5 4 = 16 + b 2 4 ⇒ 25 = 16 + b 2 ⇒ b 2 = 9 ∴ Required equation of the hyperbola is x 2 16 - y 2 9 = 1 .

If the straight line l x + m y + n = 0 be a normal to the hyperbola x 2 a 2 − y 2 b 2 = 1 , then by the application of calculus, prove that a 2 l 2 − b 2 m 2 = a 2 + b 2 2 n 2 .

Equation of hyperbola is x 2 a 2 − y 2 b 2 = 1 Equation of normal is a x sec ⁡ θ + b y tan ⁡ θ = a 2 + b 2 . . . 1 and l x + m y + n = 0 is normal to the hyperbola. ⇒ l a sec θ = m b tan θ = n − a 2 + b 2 ⇒ l sec θ a = m tan θ b = - n a 2 + b 2 ⇒ sec θ = - a n l a 2 + b 2 and tan ⁡ θ = − b n m a 2 + b 2 We know that sec 2 ⁡ θ − tan 2 ⁡ θ = 1 ⇒ a 2 n 2 l 2 a 2 + b 2 2 − b 2 n 2 m 2 a 2 + b 2 2 = 1 ⇒ a 2 n 2 l 2 − b 2 n 2 m 2 = a 2 + b 2 2 ⇒ a 2 l 2 − b 2 m 2 = a 2 + b 2 2 n 2 Hence proved.

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The equation of the transverse and conjugate axis of the hyperbola $16 x^{2} - y^{2} + 64 x + 4 y + 44 = 0$ are-

Important Questions on Hyperbola

EASY

Mathematics>Coordinate Geometry>Hyperbola>Basics of Hyperbola

The vertices of the hyperbola are at

(- 5, - 3)

and

(- 5, - 1)

and the extremities of the conjugate axis are at

(- 7, - 2)

and

(- 3, - 2),

then the equation of the hyperbola is

MEDIUM

Mathematics>Coordinate Geometry>Hyperbola>Basics of Hyperbola

The distance between the foci of a hyperbola is

16

and its eccentricity is

\sqrt{2}

. Its equation can be

HARD

Mathematics>Coordinate Geometry>Hyperbola>Basics of Hyperbola

A hyperbola, having the transverse axis of length

2 \sin θ

is confocal with the ellipse

3 x^{2} + 4 y^{2} = 12 .

Its equation is

MEDIUM

Mathematics>Coordinate Geometry>Hyperbola>Basics of Hyperbola

Find the centre, eccentricity, foci, directrices and the length of the latus rectum of the hyperbola

4 (y + 3)^{2} - 9 (x - 2)^{2} = 1

EASY

Mathematics>Coordinate Geometry>Hyperbola>Basics of Hyperbola

A hyperbola has its centre at the origin, passes through the point

(4, 2)

and has transverse axis of length

4

along the

x - a x i s .

Then the eccentricity of the hyperbola is:

MEDIUM

Mathematics>Coordinate Geometry>Hyperbola>Basics of Hyperbola

Find the eccentricity and length of the latus rectum of the hyperbola

16 y^{2} - 9 x^{2} = 144

HARD

Mathematics>Coordinate Geometry>Hyperbola>Basics of Hyperbola

A hyperbola passes through the foci of the ellipse

\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1

and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is:

MEDIUM

Mathematics>Coordinate Geometry>Hyperbola>Basics of Hyperbola

Find the foci, eccentricity, equations of the directrix, length of latus rectum of the hyperbola

x^{2} - 4 y^{2} = 4

MEDIUM

Mathematics>Coordinate Geometry>Hyperbola>Basics of Hyperbola

If a directrix of a hyperbola centered at the origin and passing through the point

(4, - 2 \sqrt{3})

5 x = 4 \sqrt{5}

and its eccentricity is

e

, then:

MEDIUM

Mathematics>Coordinate Geometry>Hyperbola>Basics of Hyperbola

The value of

b^{2}

in order that the foci of the hyperbola

\frac{x^{2}}{144} - \frac{y^{2}}{81} = \frac{1}{25}

and the ellipse

\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1

coincide is

MEDIUM

Mathematics>Coordinate Geometry>Hyperbola>Basics of Hyperbola

Find the center, foci, eccentricity, equation of directrices, length of latus rectum of the hyperbola

x^{2} - 4 y^{2} = 4

MEDIUM

Mathematics>Coordinate Geometry>Hyperbola>Basics of Hyperbola

Which of the following is the equation of a hyperbola?

MEDIUM

Mathematics>Coordinate Geometry>Hyperbola>Basics of Hyperbola

Find the equation of the hyperbola, whose foci are

(\pm 5, 0)

and length of transverse axis is

8

MEDIUM

Mathematics>Coordinate Geometry>Hyperbola>Basics of Hyperbola

Find the centre, foci, eccentricity, equation of directrices, length of the latus rectum of the hyperbola

x^{2} - 4 y^{2} = 4

MEDIUM

Mathematics>Coordinate Geometry>Hyperbola>Basics of Hyperbola

If the straight line

l x + m y + n = 0

be a normal to the hyperbola

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

, then by the application of calculus, prove that

\frac{a^{2}}{l^{2}} - \frac{b^{2}}{m^{2}} = \frac{{(a^{2} + b^{2})}^{2}}{n^{2}}

EASY

Mathematics>Coordinate Geometry>Hyperbola>Basics of Hyperbola

If the vertices and foci of a hyperbola are respectively

(\pm 3, 0)

and

(\pm 4, 0)

then the parametric equations of that hyperbola are

MEDIUM

Mathematics>Coordinate Geometry>Hyperbola>Basics of Hyperbola

An ellipse

E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

passes through the vertices of the hyperbola

H : \frac{x^{2}}{49} - \frac{y^{2}}{64} = - 1

. Let the major and minor axes of the ellipse

E

coincide with the transverse and conjugate axes of the hyperbola

H

. Let the product of the eccentricities of

E

and

H

\frac{1}{2}

. If

l

is the length of the latus rectum of the ellipse

E

, then the value of

113 l

is equal to _______.

MEDIUM

Mathematics>Coordinate Geometry>Hyperbola>Basics of Hyperbola

The equation of the directrices of the hyperbola

3 x^{2} - 3 y^{2} - 18 x + 12 y + 2 = 0

MEDIUM

Mathematics>Coordinate Geometry>Hyperbola>Basics of Hyperbola

The locus of the point of intersection of the straight lines,

t x - 2 y - 3 t = 0

and

x - 2 t y + 3 = 0 (t \in R),

is:

HARD

Mathematics>Coordinate Geometry>Hyperbola>Basics of Hyperbola

The foci of the hyperbola

16 x^{2} - 9 y^{2} - 64 x + 18 y - 90 = 0

are

The equation of the transverse and conjugate axis of the hyperbola 16x2-y2+64x+4y+44=0 are-

Important Questions on Hyperbola

Learn and Prepare for any exam you want !

The equation of the transverse and conjugate axis of the hyperbola $16 x^{2} - y^{2} + 64 x + 4 y + 44 = 0$ are-