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Visual SolutionStandard

Question

The equation of the hyperbola with vertices at

(\pm 5, 0)

and foci at

(\pm 7, 0)

is:

(A)

\frac{x^2}{25} - \frac{y^2}{24} = 1

(B)

\frac{x^2}{49} - \frac{y^2}{25} = 1

(C)

\frac{x^2}{25} - \frac{y^2}{49} = 1

(D)

\frac{x^2}{24} - \frac{y^2}{25} = 1

Solution Path

a = 5

from vertices,

c = 7

from foci,

b^2 = c^2 - a^2 = 24

. Equation:

x^2/25 - y^2/24 = 1

01Question Setup

1/4

Find the equation of the hyperbola with vertices

(\pm 5, 0)

and foci

(\pm 7, 0)

Hyperbola equation

= ;?

02Plot Hyperbola

2/4

Vertices give

a = 5

. Foci give

c = 7

. Plot both branches with asymptotes.

a = 5

c = 7

03Find b^2KEY INSIGHT

3/4

For hyperbola:

c^2 = a^2 + b^2

. So

49 = 25 + b^2

, giving

b^2 = 24

b^2 = 24

04Final Answer

4/4

Equation:

\frac{x^2}{25} - \frac{y^2}{24} = 1

\boxed{\frac{x^2}{25} - \frac{y^2}{24} = 1}

Concepts from this question2 concepts unlocked

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Eccentricity and Conic Classification

EASY

The eccentricity e = c/a determines the type of conic: e < 1 gives an ellipse, e = 1 gives a parabola, and e > 1 gives a hyperbola. For a circle, e = 0.

e = \frac{c}{a}

First step in most conic problems. JEE frequently asks you to identify or compare conics based on eccentricity values.

Conic identificationEccentricity comparisonLocus problems

Practice (12 Qs) →

Tangent Condition for Conics

STANDARD

For the line y = mx + c to be tangent to the ellipse x²/a² + y²/b² = 1, the condition is c² = a²m² + b². Similar conditions exist for parabola (c = a/m) and hyperbola (c² = a²m² - b²).

c^2 = a^2 m^2 + b^2 \quad \text{(ellipse)}

One of the most tested ideas in JEE. Lets you find tangent equations, common tangents, and locus of intersection points without calculus.

Common tangentsTangent from external pointLocus of tangent intersection

Practice (14 Qs) →

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