Hyperbola (VSAQs)

Maths-2B | 5. Hyperbola – VSAQs:
Welcome to VSAQs in Chapter 5: Hyperbola. This page contains the most important VSAQs in this chapter. Aim to secure top marks in your exams by understanding these clear and straightforward Very Short Answer Questions.

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VSAQ-1 : If e,e₁ are the eccentricities of a hyperbola and its conjugate hyperbola, then prove that 1/e² +1/(e₁²)=1

For the hyperbola $$S = \frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$$ and its conjugate hyperbola, the eccentricities are e and e₁ respectively

Then $$e = \sqrt{\frac{a^2 + b^2}{a^2}}$$ $$e_1 = \sqrt{\frac{a^2 + b^2}{b^2}}$$

$$\frac{1}{e^2} + \frac{1}{e_1^2} = \frac{a^2}{a^2 + b^2} + \frac{b^2}{a^2 + b^2} = \frac{a^2 + b^2}{a^2 + b^2} = 1$$

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VSAQ-2 : If eccentricity of a hyperbola is 5/4, then find eccentricity of its conjugate hyperbola

Let $$e = \frac{5}{4}$$ and the eccentricity of the conjugate hyperbola be e₁

Then $$\frac{1}{e^2} + \frac{1}{e_1^2} = 1$$

$$\frac{1}{(\frac{5}{4})^2} + \frac{1}{e_1^2} = 1$$

$$\frac{1}{e_1^2} = 1 – \frac{16}{25} = \frac{9}{25}$$

$$e_1^2 = \frac{25}{9}$$

$$e_1 = \frac{5}{3}$$

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VSAQ-3 : Find the angle between the asymptotes of the hyperbola x²/a² – y²/b² =1

The asymptotes of S = 0 are $$y = \pm \frac{b}{a}x$$ They are equally inclined at an angle θ to the x-axis

Here $$\tan\theta = \frac{b}{a}$$ (or) $$\theta = \tan^{-1} \frac{b}{a}$$

Now $$\sec\theta = \sqrt{1+\tan^2\theta} = \sqrt{1+\frac{b^2}{a^2}} = \sqrt{\frac{a^2+b^2}{a^2}} = e$$

$$\sec\theta = e$$

$$\theta = \sec^{-1}e$$

Hence, the angle between the 2 asymptotes is $$2\theta = 2\sec^{-1}(e)$$ (or) $$2\tan^{-1} \frac{b}{a}$$

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VSAQ-4 : If the angle between the asymptotes is 30° then find its eccentricity

The angle between the asymptotes of the hyperbola S = 0 is $$2\sec^{-1}e$$

$$2\sec^{-1}e = 300^\circ$$

$$\sec^{-1}e = 150^\circ$$

$$e = \sec150^\circ = \sqrt{6} – \sqrt{2}$$

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VSAQ-5 : Define rectangular hyperbola and find its eccentricity

Rectangular Hyperbola: In a hyperbola, if the length of the transverse axis is equal to the length of its conjugate axis, then the hyperbola is said to be rectangular hyperbola.

Thus 2a = 2b

$$a = b$$

$$e = \sqrt{\frac{a^2+b^2}{a^2}} = \sqrt{\frac{a^2+a^2}{a^2}} = \sqrt{\frac{2a^2}{a^2}} = \sqrt{2}$$

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VSAQ-6 : Find the product of lengths of the perpendiculars from any point on the hyperbola x²/16-y²/9=1 to its asymptotes

Given hyperbola $$\frac{x^2}{16} – \frac{y^2}{9} = 1$$

$$a^2 = 16 b^2 = 9$$

Product of the perpendiculars from any point on the hyperbola to its asymptotes

$$\frac{a^2b^2}{a^2+b^2} = \frac{16 \times 9}{16+9} = \frac{144}{25}$$

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VSAQ-7 : Find the equation of the hyperbola whose foci are (±5,0) the transverse axis is of length 8

Foci $$S = (±ae,0) = (±5,0)$$

ae = 5 Length of transverse axis is $$2a = 8$$

$$a = 4$$

$$b^2 = a^2(e^2-1) = (ae)^2 – a^2 = 5^2 – 4^2 = 25 – 16 = 9$$

Equation of the hyperbola is $$\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$$

$$\frac{x^2}{16} – \frac{y^2}{9} = 1$$

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VSAQ-8 : Find the value of K if 3x – 4y + k = 0 is a tangent to the hyperbola x²-4y²=5

Given hyperbola $$x^2 – 4y^2 = 5$$

$$\frac{x^2}{5} – \frac{4y^2}{5} = 1$$

$$\frac{x^2}{5} – \frac{y^2}{5/4} = 1$$

$$a^2 = 5$$ and $$b^2 = 5$$

Comparing 3x – 4y + k = 0 with lx + my + n = 0 we get $$l=3 m=-4 n=k$$

Tangential condition: $$n^2=a^2l^2-b^2m^2$$

$$k^2=5(3^2)-\frac{5}{4}(4^2) = 45-20 = 25$$

$$k^2 = 25$$

$$k = ±5$$

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VSAQ-9 : Find the equation of the normal at θ=π/3 to the hyperbola 3x²-4y²=12

Given hyperbola $$3x^2 – 4y^2 = 12$$

$$\frac{x^2}{4} – \frac{y^2}{3} = 1$$

$$a^2 = 4$$

$$a = 2 b^2 = 3$$

$$b = \sqrt{3}$$

Equation of the normal at $$p(\theta)$$ S=0 is $$ax/\sec\theta + by/\tan\theta = a^2 + b^2$$

$$2x/\sec60^\circ + \sqrt{3}y/\tan60^\circ = 4 + 3$$

$$2x/2 + \sqrt{3}y/\sqrt{3} = 7$$

$$x + y = 7$$