Solve the following Question.(1 Marks)

Question 11 Mark

If the foci of the ellipse $\frac{\text{x}^{2}}{16}+\frac{\text{y}^{2}}{\text{b}^{2}}=1$ and the hyperbola $\frac{\text{x}^{2}}{144}-\frac{\text{y}^{2}}{81}=\frac{1}{25}$ coincide, value of $\text{b}^{2}$

Answer

The given equation of the hyperbola is
$\frac{\text{x}^{2}}{144}-\frac{\text{y}^{2}}{81}=\frac{1}{25}$
$\Rightarrow\frac{25\text{x}^{2}}{144}-\frac{25\text{y}^{2}}{81}=1$
$\Rightarrow\frac{\text{x}^{2}}{\frac{144}{125}}-\frac{\text{y}^{2}}{\frac{81}{25}}=1$
$\Rightarrow\frac{\text{x}^{2}}{\Big(\frac{12}{5}\Big)^{2}}-\frac{\text{y}^{2}}{\Big(\frac{9}{5}\Big)^{2}}=1$
This equation of the hyperbola is of the form $\frac{\text{x}^{2}}{\text{a}_1^{2}}-\frac{\text{y}^{2}}{\text{b}_1^{2}}=1$
Where, $\text{a}_1^{2}=\frac{144}{25}$ and $\text{b}_1^{2}=\frac{81}{25}$
Let, e, the eccentricity of the hyperbola.Then,
$\text{e}_1=\sqrt{1+\frac{\text{b}_1^{2}}{\text{a}_1^{2}}}$
$=\sqrt{1+\frac{\frac{\frac{81}{25}}{144}}{25}}$
$=\sqrt{1+\frac{81}{144}}$
$=\sqrt{\frac{144+81}{144}}$
$=\frac{15}{12}$
$=\frac{5}{4}$
So,the coordinates of foci are $(\pm\text{a}_1\text{e}_1,0)$ i,e.$(\pm3,0)$
It is given that the foci of the elipse conicide with the foci of the hyperbola,
So, the coordinates of foci of the hyperbola are $(\pm3,0)$
Now, the given equation of ellipse is $\frac{\text{x}^{2}}{16}+\frac{\text{y}^{2}}{\text{b}^{2}}=1$
$\Rightarrow\frac{\text{x}^{2}}{4^{2}}+\frac{\text{y}^{2}}{\text{b}^{2}}=1$
This equation of the hyperbola is of the form $\frac{\text{x}^{2}}{\text{a}_2^{2}}-\frac{\text{y}^{2}}{\text{b}_2^{2}}=1$
Where, $\text{a}_2^{2}=16$ and $\text{b}_2^{2}=\text{b}^{2}$
Let,$\text{e}_2$ be the eccentricity of the given ellipse, So, the coordinates of foci are $(\pm\text{a}_2\text{e}_2, 0)$
$\therefore\text{a}_2\text{e}_2=3$
$\Rightarrow4\times\text{e}_2=3[\because\text{a}_2=4]$
$\Rightarrow\text{e}_2=\frac{3}{4}$
$\Rightarrow\text{e}_2 ^{2}=\frac{9}{16}$
Now,
$\text{b}_2^{2}=\text{a}_2^{2}(1-\text{e}_2^{2})$
$=16\Big(1-\frac{9}{16}\Big)$
$=16\times\frac{7}{16}$
$\Rightarrow\text{b}_2^{2}=7 [\because\text{b}_2^{2}=\text{b}^{2}]$
Hence,$\text{b}^{2}=7$

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Question 21 Mark

Write the lenght of the latus-rectum of the hyperbola $16\text{x}^{2}-9\text{y}^{2}=144$

Answer

The given equation of the hyperbola is
$16\text{x}^{2}-9\text{y}^{2}=144$
$\Rightarrow\frac{16\text{x}^{2}}{144}-\frac{9\text{y}^{2}}{144}=1$
$\Rightarrow\frac{\text{x}^{2}}{\frac{144}{16}}-\frac{\text{y}^{2}}{\frac{144}{9}}=1$
$\Rightarrow\frac{\text{x}^{2}}{9}-\frac{\text{y}^{2}}{16}=1$
$\Rightarrow\frac{\text{x}^{2}}{3^{2}}-\frac{\text{y}^{2}}{4^{2}}=1$
This equation is of the form $\frac{\text{x}^{2}}{\text{a}^{2}}-\frac{\text{y}^{2}}{\text{b}^{2}}=1$, where a = 3 and $\text{b}=4$
Now, the lenght of the latus-rectum = $\frac{2\text{b}^{2}}{\text{a}}$
$=2\times\frac{16}{3}$
$=\frac{32}{3}$

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Question 31 Mark

Wrie the eccentricity of the hyperbola whose latus-rectum is half of its transversw axis.

Answer

Let the equation of hyperbola be $\frac{\text{x}^{2}}{\text{a}^{2}}-\frac{\text{y}^{2}}{\text{b}^{2}}=1$
According to the question
Lenght of latus-rectum = $\frac{1}{2}$[lenght of transverse axis]
$\Rightarrow\frac{2\text{b}^{2}}{\text{a}}=\frac{1}{2}\times2\text{a}$
$\Rightarrow\frac{\text{b}^{2}}{\text{a}^{2}}=\frac{1}{2}$
Now,
$\text{e}=\sqrt{1+\frac{b^{2}}{\text{a}^{2}}}$
$=\sqrt{1+\frac{1}{2}}$
$=\sqrt{\frac{3}{2}}$
$\therefore\text{e}=\sqrt{\frac{3}{2}}$

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Question 41 Mark

If the latus-rectum throught one focus of a hyperbola subtends a right angle at the farther vertex, then write the eccentricity of the hyperbola.

Answer

It is given that ABC is right angle trianle
.$\therefore\text{AB}^{2}=\text{BC}^{2}+\text{AC}^{2}$
$\Rightarrow(\text{ae-ae})^{2}+\Big(\frac{\text{b}^{2}}{\text{a}}+\frac{\text{b}^{2}}{\text{a}}\Big)^{2}=(\text{ae-ae})^{2}$$+\Big(\frac{-\text{b}^{2}}{\text{a}}\Big)^{2}+(\text{ae+a})^{2}+\Big(\frac{\text{b}^{2}}{\text{a}}\Big)^{2}$
$\Rightarrow\frac{4\text{b}^{4}}{\text{a}^{2}}=2\text{a}^{2}(\text{e}+1)^{2}+\frac{2\text{b}^{4}}{\text{a}^{2}}$
$\Rightarrow\frac{4\text{b}^{4}-2\text{b}^{4}}{\text{a}^{2}}=2\text{a}^{2}(\text{e}+1)^{2}$
$\Rightarrow\frac{2\text{b}^{4}}{\text{a}^{2}}=2\text{a}^{2}(\text{e}+1)^{2}$
$\Rightarrow\frac{\text{b}^{4}}{\text{a}^{4}}=(\text{e}+1)^{2}---(\text{i})$
Now,
$\text{b}^{2}=\text{a}^{2}(\text{e}^{2}-1)$
$\Rightarrow\text{b}^{4}=\text{a}^{4}(\text{e}^{2}-1)^{2}$
$\Rightarrow\frac{\text{b}^{4}}{\text{a}^{4}}=(\text{e}^{2}-1{})^{2}$
$\Rightarrow(\text{e}+1)^{2}=(\text{e}^{2}-1)^{2}$
$\Rightarrow(\text{e}+1)^{2}=(\text{e}^{2}-1)(\text{e}^{2}-1)$
$\Rightarrow(\text{e}+1)^{2}=(\text{e}-1)(\text{e}+1)(\text{e}-1)$
$\Rightarrow\text{e}+1=(\text{e}-1)(\text{e}^{2}-1)$
$\Rightarrow\text{e}+1=(\text{e}-1)(\text{e}+1)$
$\Rightarrow1=(\text{e}-1)^{2}$
$1=\text{e}^{2}+1-2\text{e}$
$\Rightarrow\text{e}^{2}-2\text{e}=0$
$\text{e}-2=0$
$\Rightarrow\text{e}=2$
Hence, $\text{e}=2$

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Question 51 Mark

If e and e are respectively the eccentricities of the ellipse $\frac{\text{x}^2}{18}+\frac{\text{y}^2}{4}=1$ and the hyperbola $\frac{\text{x}^2}{9}-\frac{\text{y}^2}{4}=1,$
then write the value of $2\text{e}_1^2 +\text{e}_2^2.$

Answer

The standard from of the ellipse is $\frac{\text{x}^2}{18}+\frac{\text{y}^2}{4}=1,$ where $a^2 = 18$ and $b^2 = 4$.
So, the eccentricity is calculated in the following way:
$\Rightarrow\text{b}^2=\text{a}^2(1-\text{e}_1^2)$
$\Rightarrow4=18(1-\text{e}_1^2)$
$\Rightarrow\frac{2}{9}=1-\text{e}_1^2$
$\Rightarrow\text{e}_1^2=\frac{7}{9}$
The standard from of the hyperbola is $\frac{\text{x}^2}{9}-\frac{\text{y}^2}{4}=1,$ where $a^2 = 9$ and $b^2 = 4$.
So, the eccentricity is calculated in the following way:
$\text{b}^2=\text{a}^2(\text{e}_2^2-1)$
$\Rightarrow4=9(\text{e}_2^2-1)$
$\Rightarrow\frac{4}{9}=\text{e}_2^2-1$
$\Rightarrow\text{e}_2^2=\frac{13}{9}$
$\therefore2\text{e}_1^2+\text{e}_2^2=\frac{2\times7}{9}+\frac{13}{9}$
$=\frac{27}{9}$
$=3$

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Question 61 Mark

Write the distance between the directrices of the hyperbola $\text{x}=8\sec\theta,\ \text{y}=8\tan\theta.$

Answer

We have:
$\text{x}=8\sec\theta,\ \text{y}=8\tan\theta$
On squaring and subtracting, we get:
$\text{x}^2-\text{y}^2=64\sec^2\theta-64\tan^2\theta$
$\Rightarrow\ \text{x}^2-\text{y}^2=64$
$\Rightarrow\ \frac{\text{x}^2}{64}-\frac{\text{y}^2}{64}=1$
$\therefore$ a = b = 8
Distance between the directrices of hyperbola is $\frac{2\text{a}^2}{\sqrt{\text{a}^2+\text{b}^2}}.$
$\Rightarrow\ \frac{2\times64}{\sqrt{64+64}}$
$=\ \frac{128}{8\sqrt2}$
$=\frac{16}{\sqrt2}$
$=8\sqrt2$

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Question 71 Mark

Write the equation of the hyperbola whose vertices are $(\pm3,0)$ and foci at $(\pm5,0).$

Answer

Since, the vertices lie on x-axis, so let the equation of the required hyperbola be
$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1...(\text{i})$
The coordinates of its vertices and foci are $(\pm\text{a},0)\text{ and }(\pm\text{ae},0)$ respectively.
$\therefore\text{a} = 3[\text{vertices}:(\pm3,0)]$
$\Rightarrow\text{a} = 9$
$\text{and}, \text{ae} = 5[\because\text{foci}:(\pm5,0)]$
$\Rightarrow\ 3\times\text{e}=5$
$\Rightarrow\text{e}=\frac{5}{3}$
$\Rightarrow\text{e}^2=\frac{25}{9}$
Now,
$\text{b}^2=\text{a}^2(\text{e}^2-1)$
$\Rightarrow\text{b}^2=9\Big(\frac{25}{9}-1\Big)$
$=9\times\frac{16}{9}$
$=16$
$\Rightarrow\text{b}^2=16$
Putting $a^2 = 9$ and $b^2 = 16$ in equation $(i),$ we get
$\frac{\text{x}^2}{9}-\frac{\text{y}^2}{16}=1$
$\Rightarrow\frac{16\text{x}^2-9\text{y}^2}{144}=1$
$\Rightarrow16\text{x}^2-9\text{y}^2=144$
This is the equation of the required hyperbola.

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Question 81 Mark

Answer each of the following questions in one word or sentence or as per exact requirement of the question
write the eccentricity of the hyperbola $9\text{x}^{2}-16\text{y}^{2}=144$

Answer

We have,
$9\text{x}^{2}-16\text{y}^{2}=144$
$\Rightarrow\frac{9\text{x}^{2}}{144}-\frac{16\text{y}^{2}}{144}=1$
$\Rightarrow\frac{\text{x}^{2}}{16}-\frac{\text{y}^{2}}{9}=1$
It is of the form $\frac{\text{x}^{2}}{\text{a}^{2}}-\frac{\text{y}^{2}}{\text{b}^{2}}=1,$ where $\text{a}^{2}=16$ and $\text{b}^{2}=9$
Now,
$\text{e}=\sqrt{1+\frac{\text{b}^{2}}{\text{a}^{2}}}$
$=\sqrt{1+\frac{9}{16}}$
$=\sqrt{\frac{25}{16}}$
$=\frac{5}{4}$
Hence,e = $\frac{5}{4}$

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Question 91 Mark

Write the equation of the hyperbola of eccentricity $\sqrt{2}$ if it is known that the distance between its foci is 16

Answer

Eccentricity = e = $\sqrt{2}$
Distance between foci is
$2\text{ae}=16$
$2\text{a}\sqrt{2}=16$
$\text{a}=\frac{16}{2\sqrt{2}}=4\sqrt{2}$
$\text{e}=\frac{\sqrt{\text{a}^{2}+\text{b}^{2}}}{\text{a}}$
$\sqrt{2}=\frac{\sqrt{32+\text{b}^{2}}}{4\sqrt{2}}$
$8=\sqrt{32+\text{b}^{2}}$
$64=32+\text{b}^{2}$
$\text{b}^{2}=32$
Equation of the hyperbola is $\frac{\text{x}^{2}}{32}-\frac{\text{y}^{2}}{32}=1$
Rewriting we get, $\text{x}^{2}-\text{y}^{2}=32$

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Question 101 Mark

Wrie the coordinates of the foci of the hyperbola $9\text{x}^{2}-16\text{y}^{2}=144$

Answer

We have,
$9\text{x}^{2}-16\text{y}^{2}=144$
$\Rightarrow\frac{9\text{x}^{2}}{144}-\frac{16\text{y}^{2}}{144}=1$
$\Rightarrow\frac{\text{x}^{2}}{16}-\frac{\text{y}^{2}}{9}=1$
$\Rightarrow\frac{\text{x}^{2}}{(4)^{2}}-\frac{\text{y}^{2}}{(3)^{2}}=1$
It is the form $\frac{\text{x}^{2}}{(4)^{2}}-\frac{\text{y}^{2}}{(3)^{2}}=1$, where $\text{a}^{2}=4$ and $\text{b}^{2}=3$
Now,
$\text{e}=\sqrt{1+\frac{\text{b}^{2}}{\text{a}^{2}}}$
$=\sqrt{1+\frac{9}{16}}$
$=\frac{25}{16}$
$=\frac{5}{4}$
The coordinates of foci are $(\pm\text{ae, 0})$ i, e,.$(\pm5,0)$

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