Differentiation (Part-38) | S N De

In this article, we have solved a few Very Short Answer Type questions (9-14) of Hyperbola related problems of S N Dey mathematics class 11.

9. (i) If the length of latus rectum of a rectangular hyperbola is \(6\) unit, find its equation.

Solution

The length of the latus rectum is \(\frac{2b^2}{a} = 6 \quad \rightarrow (1)\)

For a rectangular hyperbola, \(a = b\).

So, from \((1)\) we get,

\(\frac{2a^2}{a} = 6 \Rightarrow a = 3\)

Hence, the equation of the rectangular hyperbola is given by:
\(x^2 - y^2 = a^2 \Rightarrow x^2 - y^2 = 3^2 = 9\)

(ii) Find the co-ordinates of the foci of the rectangular hyperbola \(x^2 - y^2 = 9\).

Solution

Comparing the given hyperbola \(x^2 - y^2 = 3^2\) with \(x^2 - y^2 = a^2\) we get, \(a = 3\).

Also, we know that the eccentricity \((e)\) of any rectangular hyperbola is \(\sqrt{2}\).

The co-ordinates of the foci are given by:
\((\pm ae, 0) = (\pm 3\sqrt{2}, 0)\)

10. If the latus rectum and the transverse axis of a hyperbola are equal, show that it is a rectangular hyperbola.

Solution

We know that the length of the latus rectum of the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \quad \rightarrow (1)\) is \(\frac{2b^2}{a}\) unit and the transverse axis is given by \(2a\) unit.

By question,

\(\frac{2b^2}{a} = 2a \Rightarrow b^2 = a^2\)

Hence, by \((1)\) we get,
\(\frac{x^2}{a^2} - \frac{y^2}{a^2} = 1 \Rightarrow x^2 - y^2 = a^2\)
So, the given hyperbola is a rectangular hyperbola.

11. Find the positions of the points with respect to the hyperbola \(2x^2 - y^2 = 7\).
(i) \((3, -2)\)   (ii) \((4, 5)\)   (iii) \((-2, 3)\)

Solution

We know that the point \((x_1, y_1)\) lies outside, on, or inside the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) according as \(\frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} - 1 <, = \text{ or, } > 0 \quad \rightarrow (1)\).

The given equation of the hyperbola can be written as \(\frac{x^2}{7/2} - \frac{y^2}{7} = 1\).

(i) For the point \((3, -2)\):

\(\frac{3^2}{7/2} - \frac{(-2)^2}{7} - 1 = \frac{18}{7} - \frac{4}{7} - 1 = \frac{18 - 4 - 7}{7} = 1 > 0 \quad \rightarrow (2)\)

Hence, by \((1)\) and \((2)\), the point \((3, -2)\) lies inside the given hyperbola.

(ii) For the point \((4, 5)\):

\(\frac{4^2}{7/2} - \frac{5^2}{7} - 1 = \frac{32}{7} - \frac{25}{7} - 1 = \frac{32 - 25}{7} - 1 = 1 - 1 = 0 \quad \rightarrow (2)\)

Hence, by \((1)\) and \((2)\), the point \((4, 5)\) lies on the given hyperbola. (Note: Corrected from original text which incorrectly stated 'inside' for a value of 0)

(iii) For the point \((-2, 3)\):

\(\frac{(-2)^2}{7/2} - \frac{3^2}{7} - 1 = \frac{8}{7} - \frac{9}{7} - 1 = \frac{8 - 9}{7} - 1 = -\frac{1}{7} - 1 = -\frac{8}{7} < 0 \quad \rightarrow (2)\)

Hence, by \((1)\) and \((2)\), the point \((-2, 3)\) lies outside the given hyperbola.

12. Find the position of the point \((7, 2)\) with respect to the hyperbola \(9x^2 - 16y^2 = 144\).

Solution

We know that the point \((x_1, y_1)\) lies outside, on, or inside the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) according as \(\frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} - 1 <, = \text{ or, } > 0 \quad \rightarrow (1)\).

The given equation of the hyperbola can be written as \(\frac{x^2}{16} - \frac{y^2}{9} = 1\).

Now, for the point \((7, 2)\) we get,

\(\frac{7^2}{16} - \frac{2^2}{9} - 1 = \frac{49}{16} - \frac{4}{9} - 1 = \frac{441 - 64 - 144}{144} = \frac{233}{144} > 0 \quad \rightarrow (2)\)

Hence, by \((1)\) and \((2)\), the point \((7, 2)\) lies inside the given hyperbola.

13. Show that the locus of the point of intersection of the lines \(\frac{x}{a} - \frac{y}{b} = m\) and \(m\left(\frac{x}{a} + \frac{y}{b}\right) = 1\), \(m\) being a variable parameter, is a hyperbola.

Solution

The given equations of straight lines are:

\(\frac{x}{a} - \frac{y}{b} = m \quad \rightarrow (1)\)

\(m\left(\frac{x}{a} + \frac{y}{b}\right) = 1 \quad \rightarrow (2)\)

From \((2)\), we get \(m = \frac{1}{\frac{x}{a} + \frac{y}{b}}\).

Substituting the value of \(m\) in \((1)\):

\(\frac{x}{a} - \frac{y}{b} = \frac{1}{\frac{x}{a} + \frac{y}{b}}\)

\(\Rightarrow \left(\frac{x}{a} - \frac{y}{b}\right)\left(\frac{x}{a} + \frac{y}{b}\right) = 1\)

\(\Rightarrow \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \quad \rightarrow (3)\)

Hence, by \((3)\), the locus of the point of intersection of the given lines is a hyperbola.

14. Find the parametric co-ordinates of the point \(\left(\frac{1}{\sqrt{3}}, \frac{1}{2}\right)\) on the hyperbola \(12x^2 - 4y^2 = 3\).

Solution

The equation of the hyperbola can be written as:

\(\frac{12x^2}{3} - \frac{4y^2}{3} = 1\)

\(\Rightarrow 4x^2 - \frac{4y^2}{3} = 1\)

\(\Rightarrow \frac{x^2}{(1/2)^2} - \frac{y^2}{(\sqrt{3}/2)^2} = 1 \quad \rightarrow (1)\)

Comparing \((1)\) with \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) we get,

\(a = \frac{1}{2}, \quad b = \frac{\sqrt{3}}{2}\)

So, any point on the hyperbola \((1)\) can be written in parametric form as \((a\sec\theta, b\tan\theta) \equiv \left(\frac{1}{2}\sec\theta, \frac{\sqrt{3}}{2}\tan\theta\right)\).

\(\therefore\) By question,

\(\frac{1}{2}\sec\theta = \frac{1}{\sqrt{3}}\)

\(\Rightarrow \cos\theta = \frac{\sqrt{3}}{2} = \cos 30^\circ\)

\(\Rightarrow \theta = 30^\circ\)

Hence, the parametric co-ordinates of the given point can be written as:
\(\left(\frac{1}{2}\sec 30^\circ, \frac{\sqrt{3}}{2}\tan 30^\circ\right)\)

© S N Dey Mathematics Class 11 Solutions | Hyperbola