(Each 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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