MATHS M.C.Q (1 Marks)

1. 16x² - 9y² = 144

Solution:

The vertices of the hyperbola are $(\pm3,0)$ and foci are $(\pm5,0).$

Thus, the value of a and ae are 3 and 5, respectively.

Now, using the relation b² = a²(e² - 1), we get:

b² = 25 - 9

⇒ b² = 16

Equation of hyperbola is given below:

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

16x² - 9y² = 144

3. 2e₁² + e₂² = 3

Solution:

The standard from of the ellipse is $\frac{\text{x}^2}{18}+\frac{\text{y}^2}{4}=1,$ where a² = 18 and b² = 4.

So, the eccentricity is calculated in the following way:

$\text{b}2 = \text{a}2 (1 - \text{e}_1^2)$

$\Rightarrow4 = 18 (1 - \text{e}_1^2)$

$\Rightarrow\frac{2}{9}=1-\text{e}_1^2$

$\Rightarrow\text{e}_1^2=\frac{7}{9}$

The standard from of the hyperbola is $\frac{\text{x}^2}{9}-\frac{\text{y}^2}{4}=1,$ where a² = 9 and b² = 4.

So, the eccentricity is calculated in the following way:

$\text{b}^2 = \text{a}^2(\text{e}_2^2 - 1)$

$\Rightarrow4 = 9(\text{e}_2^2 - 1)$

$\Rightarrow\frac{4}{9}=\text{e}_2^2-1$

$\Rightarrow\text{e}_2^2=\frac{13}{9}$

$\therefore2\text{e}_1^2+\text{e}_2^2=\frac{2\times7}{9}+\frac{13}{9}$

$=\frac{27}{9}$

$=3$

2. ${\frac{\sqrt5}{2}}$

Solution:

The equation of the hyperbola is x² - 4y² = 1.

This can be rewritten in the following way:

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

This is the standard form of a hyperbola, where a = 1 and $\text{b}^2=\frac{1}{4}.$

The value of eccentricity is calculated in the following way:

$\text{b}^2=\text{a}^2(\text{e}^2-1)$

$\Rightarrow\frac{1}{4}=(\text{e}^2-1)$

$\Rightarrow\text{e}^2=\frac{5}{4}$

$\Rightarrow\text{e}=\frac{\sqrt5}{4}$

1. $8\sqrt2$

Solution:

We have:

$\text{x}=8\sec\theta,\text{y}=8\tan\theta$

On squaring and subtracting:

$\text{x}^2-\text{y}^2=8\sec^2\theta-8\tan^2\theta$

$\Rightarrow\text{x}^2-\text{y}^2=8$

$\Rightarrow\frac{\text{x}^2}{8}-\frac{\text{y}^2}{8}=1$

$\therefore\text{a}=\text{b}=\text{c}$

Distance between the directrices of the hyperbola $=\frac{2\text{a}^2}{\sqrt{\text{a}^2+\text{b}^2}}$

Distance between the directrices $=\frac{2\times64}{\sqrt{64+64}}$

$=\frac{128}{8\sqrt2}$

$=\frac{16}{\sqrt2}$

$=8\sqrt2$

1. 2xy - 4x + 4y + 1 = 0

Solution:

Let P(x, y) be any point on the hyperbola.

Then, the distance of any point from the focus is eccentricity times the distance from the directrix.

$\therefore\sqrt{(\text{x}-1)^2+(\text{y}+1)^2}=\sqrt2\Big|\frac{\text{x}-\text{y}+1}{\sqrt2}\Big|$

Squaring both the sides, we get:

(x - 1)² + (y + 1)² = (x - y + 1)²

x² - 2x + 1 + y² + 1 + 2y = x² + y² + 1 - 2xy - 2y + 2x

2xy - 4x + 4y + 1 = 0

4. x² − y² = 32

Solution:

The distance between the foci is 2ae.

$\therefore$ 2ae = 16

⇒ ae = 8

$\text{e}=\sqrt2$

$\therefore\text{a}\sqrt2=8$

$\Rightarrow\text{a}=4\sqrt2$

Also, b² = a²(e² − 1)

⇒ b² = 32(2 − 1)

⇒ b² = 32

Standard form of the hyperbola is given below:

$\frac{\text{x}^2}{32}-\frac{\text{y}^2}{32}=1$

$\text{x}^2-\text{y}^2=32$

3. $\sqrt{\frac{3}{2}}$

Solution:

The lengths of the latus rectum and the transverse axis are $\frac{2\text{b}^2}{\text{a}}\text{ and }2\text{a},$ respectively.

According to the given statement, length of the latus rectum is half of its transverse axis.

$\therefore\frac{2\text{b}^2}{\text{a}}=\frac{1}{2}\times2\text{a}$

$\Rightarrow\frac{2\text{b}^2}{\text{a}}=\text{a}$

$\Rightarrow2\text{b}^2=\text{a}$

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

Substituting the value $\text{b}^2=\frac{\text{a}^2}{2},$ we get:

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

$=\frac{\text{a}\sqrt{\frac{3}{2}}}{\text{a}}$

$=\sqrt{\frac{3}{2}}$

$\therefore$ Eccentricity is $\sqrt{\frac{3}{2}}$

2. $2<\text{e}_2^2-\text{e}_1^2<3$

Solution:

The conic ​9x² + 4y² = 36 can rewritten in the following way:

$\frac{9\text{x}^2}{36}+\frac{4\text{y}^2}{36}=1$

$\Rightarrow\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}=1$

This is the standard equation of an ellipse.

$\therefore$  b² = a²(1−e₁)²

$\Rightarrow9=4(1-\text{e}_1)^2$

$\Rightarrow(\text{e}_1)^2=\frac{-5}{4}$

The conic ​9x² − 4y² = 36 can rewritten in the following way:

$\frac{9\text{x}^2}{36}-\frac{4\text{y}^2}{36}=1$

$\Rightarrow\frac{\text{x}^2}{4}-\frac{\text{y}^2}{9}=1$

This is the standard equation of a hyperbola.

$\therefore$ b² = a²(e₂² − 1)

$\Rightarrow9=4(\text{e}_2^2-1)$

$\Rightarrow(\text{e}_2)^2=\frac{13}{4}$

$\therefore\text{e}_2^2-\text{e}_1^2=\frac{13}{4}+\frac{5}{4}=2.5$

2. $\frac{\pi}{4}$

Solution:

The hyperbola $\text{x}^2 − \text{y}^2 \sec^2\alpha = 5$ can be rewritten in the following way:

$\frac{\text{x}^2}{5}-\frac{\text{y}^2}{5\cos^2\text{a}}=1$

This is the standard form of a hyperbola, where $ \text{a}^2 = 5 \text{ and } \text{b}^2 = 5\cos^2\alpha.$

$\Rightarrow\text{b}^2 = \text{a}^2(\text{e}_1^2 − 1)$

$⇒ 5\cos^2\alpha=5(\text{e}_1^2−1)$

$\Rightarrow\text{e}_1^2=\cos^2\alpha+1...(1)$

The ellipse $\text{x}^2\sec^2\alpha+\text{y}^2=25$ can be rewritten in the following way:

$\frac{\text{x}^2}{25\cos^2\alpha}+\frac{\text{y}^2}{25}=1$

This is the standard form of an ellipse, where $\text{a}^2=25\text{ and }\text{b}^2=25\cos^2\alpha$

$\text{b}^2=\text{a}^2(1-\text{e}_2^2)$

$\Rightarrow\text{e}_2^2=1-\cos^2\alpha$

$\Rightarrow\text{e}_2^2=\sin^2\alpha...(2)$

According to the question,

$\cos^2\alpha+1=3(\sin^2\alpha)$

$\Rightarrow2=4\sin^2\alpha$

$\Rightarrow\sin\alpha=\frac{1}{\sqrt2}$

$\Rightarrow\alpha=\frac{\pi}{4}$

3. $(\pm5,0)$

Solution:

The equation of the hyperbola is given below:

9x² − 16y² = 144

This equation can be rewritten in the following way:

$\frac{9\text{x}^2}{144}-\frac{16\text{y}^2}{144}=1$

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

This is the standard equation of a hyperbola, where a² = 16 and b² = 9.

The eccentricity is calculated in the following way:

b² = a²(e² − 1)

⇒ 9 = 16(e² − 1)

$\Rightarrow\frac{9}{16}=\text{e}^2-1$

$\Rightarrow\text{e}=\frac{5}{4}$

$\text{Foci}=(\pm\text{ae},0)=(\pm5,0)$

3. length of the transverse axis.

Solution:

Let P(x,y) be any point on the hyperbola, and S, S' be the focus with coordinates $(\pm\text{ae},0).$

⇒ S'P − SP = 2a

Thus, the difference of the focal distances of any point on the hyperbola is equal to the length of the transverse axis.

1. A hyperbola.

Solution:

Let P(x, y) be any point on the hyperbola $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1.$

By definition, we have:

$\text{PA}=\text{e}\big(\text{x}-\frac{\text{a}}{\text{e}}\big)=\text{ex}-\text{a}$

$\text{and }\text{PB}=\text{e}\big(\text{x}+\frac{\text{a}}{\text{e}}\big)=\text{ex}+\text{a}$

$\therefore\text{PB}−\text{PA}=\text{(ex+a)}−\text{(ex}−\text{a})=\text{2a = k}$

1. $\frac{5}{4}$

Solution:

Standard form of a hyperbola $=\frac{\text{x}^2}{16}-\frac{\text{y}^2}{9}=1$

Here, a² = 16 and y² = 9

The eccentricity is calculated in the following way:

b² = a²(e² - 1)

⇒ 9 = 16(e² - 1)

$\Rightarrow\text{e}^2-1=\frac{9}{16} $

$\Rightarrow\text{e}^2=\frac{25}{16}$

$\Rightarrow\text{e}=\frac{5}{4}$