Question

A hyperbola has its centre at the origin and passes through point (4, 2). The curve has a transverse axis of length 4 along the x axis. Find the eccentricity of the hyperbola

& \text{a)}\text{ }\dfrac{2}{\sqrt{3}} \\\ & b)\text{ }\dfrac{3}{2} \\\ & c)\text{ }\sqrt{3} \\\ & d)\text{ 2} \\\ \end{aligned}$$

Answer 1

Now for any hyperbola of the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ the eccentricity is given by $\sqrt{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}}$ . Now we can find a as 2a is the length of the transverse axis. To find b we will use the face that the hyperbola passes through (4, 2). Hence substituting x, y, a in the equation of hyperbola we can find b and hence once we have a and b we can find eccentricity of hyperbola.

Complete step-by-step answer:
Now we are given that the hyperbola has its center at origin.
We know that the equation of hyperbola which passes through is of the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$
Hence let the equation of our hyperbola be $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$
Now we know that the length of transverse axis of $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is given by 2a.
But we are given that the length of transverse axis is 4
Hence we have 2a = 4
Dividing by 2 on both sides we get
$a=2...............(1)$
Now we are given that the hyperbola passes through the point (4, 2).
Hence x = 4 and y = 2 will satisfy the equation $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$
Hence we get
$\begin{aligned} & \dfrac{{{4}^{2}}}{{{a}^{2}}}-\dfrac{{{2}^{2}}}{{{b}^{2}}}=1 \\\ & \Rightarrow \dfrac{16}{{{a}^{2}}}-\dfrac{4}{{{b}^{2}}}=1 \\\ \end{aligned}$
Now from equation (2) we get

& \Rightarrow \dfrac{16}{{{a}^{2}}}-\dfrac{4}{{{b}^{2}}}=1 \\\ & \Rightarrow \dfrac{16}{{{2}^{2}}}-\dfrac{4}{{{b}^{2}}}=1 \\\ & \Rightarrow 4-1=\dfrac{4}{{{b}^{2}}} \\\ & \Rightarrow 3{{b}^{2}}=4 \\\ & \Rightarrow {{b}^{2}}=\dfrac{4}{3}............................(2) \\\ \end{aligned}$$ Now we know that eccentricity is given by $\sqrt{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}}$ Hence from (1) and equation (2) we get eccentricity of hyperbola is $$\begin{aligned} & \sqrt{1+\dfrac{\dfrac{4}{3}}{{{2}^{2}}}} \\\ & =\sqrt{1+\dfrac{1}{3}} \\\ & =\sqrt{\dfrac{4}{3}} \\\ & =\dfrac{2}{\sqrt{3}} \\\ \end{aligned}$$ Hence the eccentricity of ellipse is $\dfrac{2}{\sqrt{3}}$ **So, the correct answer is “Option A”.** **Note:** Now equation of hyperbola is $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ the eccentricity is given by $\sqrt{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}}$ . And the equation of ellipse is $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ and its eccentricity is $\sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}}$