Question

Find the equation of hyperbola (with a diagram) that satisfies the given conditions: foci $\left( \pm 3\sqrt{5},0 \right)$ & length of latus rectum is 8 units. Find the eccentricity of the above hyperbola. Find if $\left( -1,2 \right)$ lie on the hyperbola or not?

Answer 1

We are given a hyperbola whose foci is $\left( \pm 3\sqrt{5},0 \right)$ and length of latus rectum is 8 units. Since the x-coordinate of foci is 0, therefore the focus lies on the x-axis. Hence, the equation of hyperbola will be in form of: $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, where a > b. So, we need to identify the values of a and b for getting the equation of hyperbola. As we know that foci is represented as $\left( \pm ae,0 \right)$, where e is the eccentricity given by $e=\dfrac{\sqrt{{{a}^{2}}+{{b}^{2}}}}{a}$. Also, the length of the latus rectum for hyperbola is given by: $\dfrac{2{{b}^{2}}}{a}$. Get the value of b in terms of a and put in the eccentricity to get a quadratic equation of a. Now, solve the equation to get the value of a and simultaneously get b and e. Hence, put the values of a and b to get the equation of hyperbola.
To check if the point $\left( -1,2 \right)$, put in the equation of hyperbola and see if the LHS = RHS.

Complete step-by-step answer:
We have a hyperbola whose foci is $\left( \pm 3\sqrt{5},0 \right)$ and the length of the latus rectum is 8 units.
As we know that length of latus rectum of hyperbola is given by $\dfrac{2{{b}^{2}}}{a}$
So, we can write:
$\begin{aligned} & \dfrac{2{{b}^{2}}}{a}=8 \\\ & {{b}^{2}}=4a......(1) \\\ \end{aligned}$
Also, we have, eccentricity as: $e=\dfrac{\sqrt{{{a}^{2}}+{{b}^{2}}}}{a}$
So, we can write $ae=\sqrt{{{a}^{2}}+{{b}^{2}}}......(2)$
From the focii of the hyperbola, we have: $ae=3\sqrt{5}$
So, we can write equation (2) as:
$\sqrt{{{a}^{2}}+{{b}^{2}}}=3\sqrt{5}......(3)$
Now, square both sides of equation (3), we get:
${{a}^{2}}+{{b}^{2}}=45......(4)$
As we know that, ${{b}^{2}}=4a$ from equation (1), we can write equation (4) as:

& {{a}^{2}}+4a=45 \\\ & {{a}^{2}}+4a-45=0......(5) \\\ \end{aligned}$$ Now, solve the quadratic equation, we get: $\begin{aligned} & a\left( a+9 \right)-5\left( a+9 \right)=0 \\\ & \left( a+9 \right)\left( a-5 \right)=0 \\\ & a=-9,5 \\\ \end{aligned}$ Since value of a is non-negative, Therefore a = 5 So, we get b = 20 Hence the equation of hyperbola is $$\dfrac{{{x}^{2}}}{25}-\dfrac{{{y}^{2}}}{20}=1$$ The hyperbola we get is:  Now, to check if the point $\left( -1,2 \right)$, put in the equation of hyperbola, put x = -1 and y = 2 in equation of hyperbola, we get: $$\begin{aligned} & LHS=\dfrac{{{(-1)}^{2}}}{25}-\dfrac{{{(2)}^{2}}}{20} \\\ & =\dfrac{1}{25}-\dfrac{4}{20} \\\ & =\dfrac{-1}{100} \end{aligned}$$ Hence, $LHS\ne RHS$ Therefore, the point $\left( -1,2 \right)$ does not lie on the hyperbola. **Note:** Firstly, identify which hyperbola is given in the question. It makes it easier to write the equation of the hyperbola. Be careful while working with the formula of eccentricity, foci and latus rectum of the hyperbola.