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This is the second of our many quizzes on Higher order derivatives. Click on this to access more such competitive quizzes on Calculus.

Higher order derivatives Quiz 2

mathematics

 A hyperbola whose foci are $\left( { \pm 35,0} \right)$ and length of transverse axis is $50$ is:

 First find the value of ${b^2}$, then apply equation of hyperbola $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$.

$  foci = \left( { \pm 35,0} \right) \\  {\text{Transver axis is }}50 \\   \Rightarrow 2a = 50 \\   \Rightarrow a = 25 \\  {a^2} + {b^2} = {c^2} \\   \Rightarrow {\left( {25} \right)^2} + {b^2} = {\left( {35} \right)^2} \\   \Rightarrow {b^2} = 600 \\   \Rightarrow {\text{Equation of hyperbola is }} = \dfrac{{{x^2}}}{{625}} + \dfrac{{{y^2}}}{{600}} = 1$

 Determine the equation of hyperbola having vertices $\left( {0, \pm 6} \right)$ and has foci $\left( {0, \pm 13} \right)$.

 Use coordinates of vertices and foci to find value of $a$and $b$. Then substitute those values in the equation of hyperbola.

${\text{Consider the vertices of hyperbola}}{\text{.}} \\  \left( {0, \pm a} \right) = \left( {0, \pm 6} \right) \\   \Rightarrow a = 6 \\  {\text{Consider the foci of hyperbola}} \\  \left( {0, \pm c} \right) = \left( {0, \pm 13} \right) \Rightarrow c = 13 \\  {\text{Further solve,}} \\  \sqrt {{a^2} + {b^2}}  = 13 \Rightarrow {a^2} + {b^2} = 169 \\   \Rightarrow {b^2} = 169 - 36 = 133 \\  {\text{Consider the equation of hyperbola}} \\  \dfrac{{{y^2}}}{{{a^2}}} - \dfrac{{{x^2}}}{{{b^2}}} = 1 \\   \Rightarrow \dfrac{{{y^2}}}{{36}} - \dfrac{{{x^2}}}{{133}} = 1 \\  $

 Equation of hyperbola is $\dfrac{{{x^2}}}{{317}} - \dfrac{{{y^2}}}{{219}} = 1$. Let $a$ be the $x$ coordinate of the vertices of hyperbola and  be the length of its transverse axis. What is $a + \ell $

 Use $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ to find the value of  and further apply length of transverse.

$  {\text{Given equation,}} \\  \dfrac{{{x^2}}}{{317}} - \dfrac{{{y^2}}}{{219}} = 1 \\  {\text{Length of transverse axis}} = 2a \\  \therefore a = \sqrt {317}  \\   \Rightarrow \ell  = 2 \times \sqrt {317}  \\   \Rightarrow \ell  = 2\sqrt {317}  \\   \Rightarrow a + \ell  = 3\sqrt {317}  \\  $

 What are the coordinates of vertices and foci of the hyperbola having equation $\dfrac{{{x^2}}}{{49}} - \dfrac{{{y^2}}}{9} = 1$

 Compare the given equation of hyperbola with the general equation of hyperbola and get $a$ and $b$. Then substitute these values in formula for coordinates of vertices, foci.

$  {\text{Consider the given equation of hyperbola}}{\text{.}} \\  \dfrac{{{x^2}}}{{49}} - \dfrac{{{y^2}}}{9} = 1\,\,.....(i) \\  {\text{Consider the general equation of hyperbola}}{\text{.}} \\  \dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\,\,.....(ii) \\  {\text{Compare equations }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \\  {a^2} = 49 \Rightarrow a =  \pm 7 \\  {b^2} = 9 \Rightarrow b =  \pm 3 \\  {\text{Vertices  = }}\left( { \pm a,0} \right) = \left( { \pm 7,0} \right) \\  c = \sqrt {{a^2} + {b^2}}  \\   \Rightarrow c = \sqrt {49 + 9}  = \sqrt {58}  \\  foci = \left( { \pm c,0} \right) = \left( { \pm \sqrt {58} ,0} \right) \\  $

 Derive the equation of hyperbola whose length of latus rectum is $\dfrac{9}{2}$ and the eccentricity is $\dfrac{5}{4}$. 

 Use the given conditions to find $a$and $b$, and then substitute them in the equation of hyperbola.

$  {\text{Consider the eccentricity of the hyperbola}}{\text{.}} \\  e = \dfrac{c}{a} = \dfrac{{\sqrt {{a^2} + {b^2}} }}{a} = \dfrac{5}{4}\,\,\,...(i) \\  {\text{Consider the length of latus rectum of hyperbola}}{\text{.}} \\  \dfrac{{2{b^2}}}{a} = \dfrac{9}{2} \Rightarrow {b^2} = \dfrac{{9a}}{4}\,\,\,....(ii) \\  {\text{Substitute value of }}\,{b^2}{\text{ from equation }}\left( {{\text{ii}}} \right){\text{ in equation }}\left( {\text{i}} \right) \\  \dfrac{{\sqrt {{a^2} + \dfrac{{9a}}{4}} }}{a} = \dfrac{5}{4} \Rightarrow {a^2} + \dfrac{{9a}}{4} = \dfrac{{25{a^2}}}{{16}} \\   \Rightarrow \dfrac{{9{a^2}}}{{16}} = \dfrac{{9a}}{4} \\   \Rightarrow a = 4 \Rightarrow {a^2} = 16 \\  {\text{Substitute }}a = 4{\text{ in equation }}\left( {{\text{ii}}} \right) \\  {b^2} = \dfrac{{9 \times 4}}{4} = 9 \\  {\text{Consider the equation of hyperbola}} \\  \dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1 \\   \Rightarrow \dfrac{{{x^2}}}{{16}} - \dfrac{{{y^2}}}{9} = 1 \\  $