Q. 37 - Calculus

Question

Use Cartesian coordinates to express the equations for the ellipses determined by the conditions specified in Exercises 32–37. 

$\text{ foci }(0,\pm \alpha )\text{, minor axis }2\alpha$

Step-by-Step Solution

Answer

The answer is $\frac{x^2}{{\alpha }^2}+\frac{y^2}{2{\alpha }^2}=1\text{. }$

Step 1. Given information.

The given values are,

$\text{ foci }(0,\pm \alpha )\text{, minor axis }2\alpha$.

Step 2. values of the variables.

The foci are $(0,\alpha )(0,-\alpha )$

Centre$=\left(\frac{0+0}{2},\frac{\alpha -\alpha }{2}\right)\left[\text{ since mid point }=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\right]$$=(0,0)$

$$\begin{gathered} \text{ Given minor axis is }2\alpha \text{. } \\ \text{ Then }2b=2\alpha \\ b=\alpha \\ \left.c=\sqrt{(0-0{)}^2+(0-\alpha {)}^2} \text{ since }D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\right] \\ \text{ That is the distance from }(0,\alpha )(0,0)\text{. } \\ c=\alpha \\ \text{ Now take }{c}^2=a^2-b^2 \\ {c}^2=a^2-b^2 \\ {\alpha }^2=a^2-{\alpha }^2 \\ {a}^2=2{\alpha }^2 \end{gathered}$$

Step 3. Substitution.

On substituting,

$$\begin{gathered} \frac{(x-h{)}^2}{b^2}+\frac{(y-k{)}^2}{a^2}=1\text{ where }a>b,(h,k)\text{ Is the center. } \\ \frac{(x-0{)}^2}{{\alpha }^2}+\frac{(y-0{)}^2}{2{\alpha }^2}=1 \\ \frac{x^2}{{\alpha }^2}+\frac{y^2}{2{\alpha }^2}=1\text{. } \end{gathered}$$