Differentiation (Part-38) | S N De

In this article, we have solved a few Very Short Answer Type questions (1–8) of Hyperbola related problems from S.N. Dey Mathematics, Class 11.

1. Explain what general diagrams are represented by the equation $\frac{x^2}{a^2} + \frac{y^2}{a^2(1-e^2)} = 1$, where $a$ and $e$ are constants.

Solution

We have the equation given by:

$$ \frac{x^2}{a^2} + \frac{y^2}{a^2(1-e^2)} = 1 \quad \text{--- (1)} $$

There can be three conditions for different values of $e$:

Case (a): $0 < e < 1$

For $0 < e < 1$, let $1-e^2 = c$ (where $c > 0$). In this case, we get from (1):

$$ \frac{x^2}{a^2} + \frac{y^2}{a^2 \cdot c} = 1 $$ $$ \text{or,} \quad \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad \text{(where } b^2 = a^2 \cdot c \text{)} \quad \text{--- (2)} $$

Clearly, (2) represents the equation of an ellipse.

Case (b): $e = 0$

For $e = 0$, we get from (1):

$$ \frac{x^2}{a^2} + \frac{y^2}{a^2(1-0)} = 1 $$ $$ \text{or,} \quad x^2 + y^2 = a^2 \quad \text{--- (3)} $$

Equation (3) represents a circle.

Case (c): $e > 1$

For $e > 1$, $1-e^2 < 0$. Let $1-e^2 = -c$ (where $c > 0$). So, from (1) we get:

$$ \frac{x^2}{a^2} + \frac{y^2}{a^2(-c)} = 1 $$ $$ \text{or,} \quad \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \quad \text{(where } b^2 = a^2 c \text{)} \quad \text{--- (4)} $$

Equation (4) represents a hyperbola.

2. Show that the eccentricity of any rectangular hyperbola is $\sqrt{2}$.

Solution

We know that the general form of a hyperbola is:

$$ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \quad \text{--- (1)} $$

The eccentricity ($e$) of (1) is given by:

$$ e = \sqrt{1 + \frac{b^2}{a^2}} \quad \text{--- (2)} $$

For a rectangular hyperbola, $a = b$. Substituting this into (2), we get:

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

Hence proved.

3. Find the eccentricity, coordinates of the foci, and the equations of the directrices of the hyperbolas:
(i) $16x^2 - 9y^2 = 144$    (ii) $4x^2 - 9y^2 = 36$

Solution (i) $$ 16x^2 - 9y^2 = 144 $$ $$ \text{or,} \quad \frac{x^2}{9} - \frac{y^2}{16} = 1 \quad \text{--- (1)} $$

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

$a^2 = 9 \Rightarrow a = 3$,   $b^2 = 16 \Rightarrow b = 4$.

$$ \therefore e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3} $$

Coordinates of foci: $(\pm ae, 0) = \left(\pm 3 \times \frac{5}{3}, 0\right) = (\pm 5, 0)$

Equations of directrices:

$$ x = \pm \frac{a}{e} = \pm \frac{3}{\frac{5}{3}} = \pm \frac{9}{5} $$ $$ \text{or,} \quad 5x = \pm 9 $$

---

Solution (ii) $$ 4x^2 - 9y^2 = 36 $$ $$ \text{or,} \quad \frac{x^2}{9} - \frac{y^2}{4} = 1 \quad \text{--- (1)} $$

Comparing (1) with the general form, we get:

$a^2 = 9 \Rightarrow a = 3$,   $b^2 = 4 \Rightarrow b = 2$.

$$ \therefore e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3} $$

Coordinates of foci: $(\pm ae, 0) = \left(\pm 3 \times \frac{\sqrt{13}}{3}, 0\right) = (\pm \sqrt{13}, 0)$

Equations of directrices:

$$ x = \pm \frac{a}{e} = \pm \frac{3}{\frac{\sqrt{13}}{3}} = \pm \frac{9}{\sqrt{13}} $$ $$ \text{or,} \quad \sqrt{13}x = \pm 9 $$

4. Show that the eccentricities of the two hyperbolas $\frac{x^2}{16} - \frac{y^2}{9} = 1$ and $\frac{x^2}{64} - \frac{y^2}{36} = 1$ are equal.

Solution

We have two hyperbolas given by:

$$ \frac{x^2}{16} - \frac{y^2}{9} = 1 \quad \text{--- (1)} $$ $$ \frac{x^2}{64} - \frac{y^2}{36} = 1 \quad \text{--- (2)} $$

Comparing (1) with the general form, we get $a^2 = 16, b^2 = 9$:

$$ \therefore e_1 = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4} $$

Comparing (2) with the general form, we get $a^2 = 64, b^2 = 36$:

$$ \therefore e_2 = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{36}{64}} = \sqrt{\frac{64 + 36}{64}} = \sqrt{\frac{100}{64}} $$ $$ \text{or,} \quad e_2 = \frac{10}{8} = \frac{5}{4} $$

$$ \therefore e_1 = e_2 $$

(Where $e_1$ and $e_2$ are the eccentricities of the first and second hyperbola, respectively.)

5(i) Find the coordinates of the foci of the hyperbola $x^2 - y^2 + 1 = 0$.

Solution

Given equation:

$$ x^2 - y^2 + 1 = 0 $$ $$ \implies \frac{y^2}{1} - \frac{x^2}{1} = 1 \quad \text{--- (1)} $$

Comparing (1) with the general form of a hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, we get:

$$ a^2 = 1 \implies a = 1, \quad b^2 = 1 $$ $$ \therefore e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{1}{1}} = \sqrt{2} $$

The coordinates of the foci of the hyperbola (1) are:

$$ (0, \pm ae) = (0, \pm 1 \times \sqrt{2}) = (0, \pm \sqrt{2}) $$

5(ii) Find the eccentricity and the length of the latus rectum of the hyperbola $x^2 - y^2 = 2$.

Solution

Given equation:

$$ x^2 - y^2 = 2 \implies \frac{x^2}{2} - \frac{y^2}{2} = 1 \quad \text{--- (1)} $$

Comparing (1) with the general form of a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we get:

$$ a^2 = 2, \quad b^2 = 2 \implies a = b = \sqrt{2} $$

So, the eccentricity is:

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

The length of the latus rectum is:

$$ \frac{2b^2}{a} = \frac{2 \times 2}{\sqrt{2}} = 2\sqrt{2} \text{ units} $$

6. Find the length of the latus rectum and the equations of the directrices of the hyperbola $3y^2 - 4x^2 = 12$.

Solution

Given equation:

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

Comparing (1) with the general form of a hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, we get:

$$ a^2 = 4 \implies a = 2, \quad b^2 = 3 $$

The length of the latus rectum of the hyperbola (1) is:

$$ \frac{2b^2}{a} = \frac{2 \times 3}{2} = 3 \text{ units} $$

To find the equations of the directrices, we first calculate the eccentricity $e$:

$$ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{3}{4}} = \sqrt{\frac{7}{4}} = \frac{\sqrt{7}}{2} $$

Since the transverse axis is along the y-axis, the equations of the directrices are $y = \pm \frac{a}{e}$:

$$ y = \pm \frac{2}{\frac{\sqrt{7}}{2}} = \pm \frac{4}{\sqrt{7}} $$ $$ \implies \sqrt{7}y = \pm 4 $$

7. What type of conic is represented by the equation $x^2 - y^2 = 4$? What is its eccentricity?

Solution

The given equation can be rewritten in standard form as:

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

Since $a^2 = b^2 = 4$, this represents a rectangular hyperbola.

Its eccentricity is:

$$ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{4}{4}} = \sqrt{1 + 1} = \sqrt{2} $$

8. If the length of the conjugate axis and the length of the latus rectum of a hyperbola are equal, find its eccentricity.

Solution

For the standard hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$:

• Length of the conjugate axis = $2b$ units
• Length of the latus rectum = $\frac{2b^2}{a}$ units

According to the question:

$$ 2b = \frac{2b^2}{a} $$ $$ \implies \frac{b}{a} = 1 $$

Therefore, the eccentricity is:

$$ e = \sqrt{1 + \left(\frac{b}{a}\right)^2} = \sqrt{1 + 1^2} = \sqrt{1 + 1} = \sqrt{2} $$