Question

The eccentricity of the hyperbola x^2/a^2 - y^2/b^2 = 1 which passes through the points (3,0) and(3√(2 ,2)is ______________ *​

Answer 1

EXPLANATION.

Eccentricity of the hyperbola,

⇒ x²/a² - y²/b² = 1.

Passes through the point = (3,0) & (3√2,2).

As we know that,

(3,0) passes through the hyperbola,

put the point in the hyperbola, we get.

⇒ (3)²/a² - (0)²/b² = 1.

⇒ 9/a² - 0 = 1.

⇒ 9 = a².

Now, equation is written as,

⇒ x²/9 - y²/b² = 1.

(3√2,2) also passes trough the point of hyperbola,

put the point in the hyperbola, we get.

⇒ (3√2)²/9 - (2)²/b² = 1.

⇒ 18/9 - 4/b² = 1.

⇒ 2 - 4/b² = 1.

⇒ -4/b² = 1 - 2.

⇒ -4/b² = -1.

⇒ 4/b² = 1.

⇒ b² = 4.

Now, equation is written as,

⇒ x²/9 - y²/4 = 1.

As we know that,

Eccentricity of a hyperbola,

⇒ b² = a²(e² - 1).

⇒ 4 = 9(e² - 1).

⇒ 4 = 9e² - 9.

⇒ 4 + 9 = 9e².

⇒ 13 = 9e².

⇒ 13/9 = e².

⇒ e = √13/3.

                                                                                       

MORE INFORMATION.

(1) = Pair of tangents = SS₁ = T².

(2) = Chord of contact = T = 0 at (x₁, y₁).

(3) = Equation of the chord whose mid point is given T = S₁.

(4) = Director Circle = The equation of the director circle is x² + y² = a² - b² (a² > b²).

Answer 2

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