6 Most SAQ’s of Hyperbola Chapter in Inter 2nd Year Maths-2B (TS/AP)

SAQ-1 : Find the centre, eccentricity, foci, length of latus rectum and equations of the directrices of the hyperbola x²-4y²=4

Given hyperbola is $$x^2 – 4y^2 = 4$$

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

Here $$a^2 = 4 b^2 = 1$$

(i) Centre C = (0,0)

(ii) Eccentricity $$e = \sqrt{\frac{a^2+b^2}{a^2}} = \sqrt{\frac{4+1}{4}} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}$$

(iii) Foci = $$(\pm ae,0)$$ $$= (\pm 2\left(\frac{\sqrt{5}}{2}\right),0) = (\pm \sqrt{5},0)$$

(iv) Equation of the directrices is $$x = \pm \frac{a}{e}$$

$$\Rightarrow x = \pm \frac{2}{\sqrt{5}/2}$$

$$\Rightarrow x = \pm \frac{4}{\sqrt{5}}$$

(v) Length of the latus rectum = $$2b^2/a = 2(1)/2 = 1$$

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SAQ-2 : Find the centre, eccentricity, foci, length of latus rectum and equations of the directrices of the hyperbola 16y²-9x²=144

Given hyperbola is $$16y^2 – 9x^2 = 144$$

$$\Rightarrow \frac{16y^2}{144} – \frac{9x^2}{144} = 1$$

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

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

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

(i) Centre C = (0,0)

(ii) $$e = \sqrt{\frac{a^2 + b^2}{b^2}} = \sqrt{\frac{16 + 9}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}$$

(iii) Foci = $$(0, \pm be) = (0, \pm 3 \cdot \frac{5}{3}) = (0, \pm 5)$$

(iv) Equation of the directrices is $$y = \pm \frac{b}{e}$$

$$\Rightarrow y = \pm \frac{3}{\frac{5}{3}}$$

$$\Rightarrow y = \pm \frac{9}{5}$$

(v) Length of latus rectum = $$\frac{2a^2}{b} = \frac{2 \cdot 16}{3} = \frac{32}{3}$$

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SAQ-3 : Find the equation of the tangents to the hyperbola x²-4y²=4 which are
i.Parallel
ii.Perpendicular to the line x + 2y =0

Given hyperbola is $$x^2 – 4y^2 = 4$$

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

Here $$a^2 = 4 b^2 = 1$$

Slope of the given line x + 2y = 0 is $$m = -\frac{1}{2}$$

$$\Rightarrow \text{Slope of its perpendicular is } 2$$

Formula for the tangent with slope m to a hyperbola $$\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$$

$$y = mx \pm \sqrt{a^2m^2 – b^2}$$

(i) Parallel tangent with slope $$-\frac{1}{2}$$ is $$y = -\frac{1}{2}x \pm \sqrt{4\left(\frac{1}{4}\right) – 1}$$

This simplifies incorrectly in the provided steps; let’s correct that:

$$y = -\frac{1}{2}x \pm \sqrt{4\left(\frac{1}{4}\right) – 1} = -\frac{1}{2}x \pm \sqrt{1 – 1} = -\frac{1}{2}x$$

$$x + 2y = 0$$

(ii) Perpendicular tangent with slope 2 is:

$$y = 2x \pm \sqrt{4(2^2) – 1}$$

$$\Rightarrow y = 2x \pm \sqrt{16 – 1} = 2x \pm \sqrt{15}$$

So, the perpendicular tangent with slope 2 is $$y = 2x \pm \sqrt{15}$$

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SAQ-4 : Find the equations of the tangents to the hyperbola 3x²-4y²=12 which are
a.Parallel
b.Perpendicular to the line y = x – 7

Given hyperbola is $$3x^2 – 4y^2 = 12$$

$$\Rightarrow \frac{x^2}{4} – \frac{y^2}{3} = 1$$

Here $$a^2 = 4 b^2 = 3$$

The slope of the given line y = x – 7 is m = 1

$$\Rightarrow \text{Slope of its perpendicular is } -1$$

Formula: Tangent with slope m is $$y = mx \pm \sqrt{a^2m^2 – b^2}$$

(i) Parallel tangent with slope m = 1 is

$$y = x \pm \sqrt{4(1)^2 – 3} = x \pm \sqrt{4 – 3} = x \pm 1$$

$$\Rightarrow x – y \pm 1 = 0$$

(ii) Perpendicular tangent with slope m = -1 is

$$y = (-1)x \pm \sqrt{4(1)^2 – 3} = -x \pm \sqrt{1}$$

$$\Rightarrow x + y \pm 1 = 0$$

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SAQ-5 : Prove that the point of intersection of two perpendicular tangents to the hyperbola x²/a² -y²/b²-1=0 lies on the circle x²+y²=a²-b²

Given hyperbola is $$S = \frac{x^2}{a^2} – \frac{y^2}{b^2} – 1 = 0$$

Let $$P(x_1,y_1)$$ be a point on the locus

A tangent with slope m is $$y = mx \pm \sqrt{a^2m^2 – b^2}$$

If it passes through $$P(x_1,y_1)$$ then $$y_1 – mx_1 = \pm \sqrt{a^2m^2 – b^2}$$

$$\Rightarrow (y_1 – mx_1)^2 = a^2m^2 – b^2$$

$$\Rightarrow (y_1^2 – 2mx_1y_1 + m^2x_1^2) – a^2m^2 + b^2 = 0$$

$$\Rightarrow m^2(x_1^2 – a^2) – 2mx_1y_1 + (y_1^2 + b^2) = 0 \quad …(1)$$

Equation (1) is a quadratic equation in m, and its roots are taken as $$m_1, m_2$$

If the tangents are perpendicular, then $$m_1m_2 = -1$$

From (1), the product of the roots $$\frac{y_1^2 + b^2}{x_1^2 – a^2} = -1$$

$$\Rightarrow x_1^2 + y_1^2 = a^2 – b^2$$

The locus of $$P(x_1,y_1)$$ is $$x^2 + y^2 = a^2 – b^2$$

This is called the director circle.

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SAQ-6 : Tangents to the hyperbola x²/a² -y²/b² =1 make angles θ₁,θ₂ with transverse axis of a hyperbola. Show that the point of intersection of these tangents lies on the curve 2xy = k(x²-a²) when tanθ₁+tanθ₂=k

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

Let $$P(x_1,y_1)$$ be a point on the locus.

A tangent with slope m is $$y = mx \pm \sqrt{a^2m^2 – b^2}$$

If it passes through $$P(x_1,y_1)$$ then $$y_1 = mx_1 \pm \sqrt{a^2m^2 – b^2}$$

$$\Rightarrow y_1 – mx_1 = \pm \sqrt{a^2m^2 – b^2}$$

$$\Rightarrow (y_1 – mx_1)^2 = a^2m^2 – b^2$$

$$\Rightarrow (y_1^2 – 2mx_1y_1 + m^2x_1^2) – a^2m^2 + b^2 = 0$$

$$\Rightarrow m^2(x_1^2 – a^2) – 2mx_1y_1 + (y_1^2 + b^2) = 0 \quad …(1)$$

Equation (1) is a quadratic in m, and its roots are taken as $$m_1, m_2$$

From (1), the sum of the roots $$m_1 + m_2 = \frac{2x_1y_1}{x_1^2 – a^2}$$

$$\Rightarrow \tan\theta_1 + \tan\theta_2 = \frac{2x_1y_1}{x_1^2 – a^2}$$

$$\Rightarrow k = \frac{2x_1y_1}{x_1^2 – a^2}$$

$$\Rightarrow 2x_1y_1 = k(x_1^2 – a^2)$$

The equation of the locus of $$P(x_1,y_1)$$ is $$2xy = k(x^2 – a^2)$$