Test Index

Hyperbola

Section: Mathematics

Question : 1 of 34

Marks: +1, -0

3 \sqrt{2} x - 4 y = 12

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

\; \frac{5}{4}

a^{2} - b^{2} =

Solution:

Given, equation of hyperbola is

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

Now, the equation of the tangent to be hyperbola is

\; 3 \sqrt{2} x - 4 y = 12

\; \Rightarrow \; \; y = \; \frac{3 \sqrt{2}}{4} x - 3

So, slope of the tangent,

m_t = \; \frac{3 \sqrt{2}}{4}

and

y

-intercept is

c = -3

Now, the line

y = m x + c

tangent to the hyperbola is given by

c^{2} \; = a^{2} m^{2} - b^{2}

\Rightarrow \; \; (-3)^{2} \; = a^{2} \left( \frac{3 \sqrt{2}}{4} \right)^{2} - b^{2}

\Rightarrow \; \; 9 \; = \; \frac{9}{8} a^{2} - b^{2}

\Rightarrow \; \; 9 a^{2} - 8 b^{2} \; = 72

Given, eccentricity,

e = \; \frac{5}{4}

\; \text{So,} \; b^{2} \; = a^{2} (e^{2} - 1)

\; = a^{2} \left( \left( \frac{5}{4} \right)^{2} - 1 \right)

\; = a^{2} \left( \frac{25}{16} - 1 \right) = \frac{9}{16} a^{2}

\Rightarrow \; \; 16 b^{2} \; = 9 a^{2}

From Eqs. (i) and (ii), we get

b^{2} \; = 9 \; \text{and} \; a^{2} = 16

\therefore a^{2} - b^{2} \; = 16 - 9 = 7

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