4.0 Difference between two forms of Ellipse

• If the centre of ellipse is $(\alpha ,\beta )$ instead of origin $(0,0)$ then equation of ellipse is $$\frac{{{{(x - \alpha )}^2}}}{{{a^2}}} + \frac{{{{(y - \beta )}^2}}}{{{b^2}}} = 1$$
• Generally, the equation of ellipse, whose focus is $(h,k)$ and directrix is $lx + my + n = 0$ and eccentricity is $e$, is $${(x - h)^2} + {(y - k)^2} = {e^2} \times \frac{{{{(lx + my + n)}^2}}}{{{l^2} + {m^2}}}$$

Question 2. The equation of ellipse is ${x^2} + 4{y^2} + 2x + 16y + 13 = 0$. Find the eccentricity, vertex and focus for the given ellipse.

Solution: We can write the equation of ellipse as

$$\begin{equation} \begin{aligned} {x^2} + 2x + 1 - 1 + 4{y^2} + 16y + 16 - 16 + 13 = 0 \\ {(x + 1)^2} + {(2y + 4)^2} = 4 \\ {(x + 1)^2} + 4{(y + 2)^2} = 4 \\ \frac{{{{(x + 1)}^2}}}{4} + {(y + 2)^2} = 1 \\\end{aligned} \end{equation} $$

On comparing it with the standard equation of ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, we get ${a^2} = 4$ and ${b^2} = 1$ or, $a=2(a>b)$ and $b=1$

Put $X=x+1$ and $Y=y+2$, the equation of ellipse becomes $$\frac{{{X^2}}}{4} + \frac{{{Y^2}}}{1} = 1...(1)$$Now, the eccentricity of ellipse $e = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} $

$$e = \sqrt {1 - \frac{1}{4}} = \sqrt {\frac{3}{4}} = \frac{{\sqrt 3 }}{2}$$

and

The vertex of ellipse given in equation $(1)$ is $( \pm a,0)$ i.e., $( \pm 2,0)$.

But the Vertex of ellipse given in question can be find out by the following equations, since the centre of ellipse is not origin.

$ \pm 2 = x + 1$ and $0 = y + 2$

$x = 1$ or $x=-3$ and $y=-2$

Therefore, the vertex of the ellipse given in question are $(1, - 2)$ and $( - 3, - 2)$.

and

Similarly, Focus of ellipse given in equation $(1)$ is $( \pm ae,0)=( \pm \sqrt 3 ,0)$.

The focus of the ellipse given in question is

$ \pm \sqrt 3 = x + 1$ and $0 = y + 2$

$x = \sqrt 3 - 1$ or $x = - \sqrt 3 - 1$ and $y = - 2$

Therefore, the focus of ellipse are $(\sqrt 3 - 1, - 2)$ and $( - \sqrt 3 - 1, - 2)$.

Question 3. Find the equation of ellipse in standard form whose length of major axis is $26$ and foci $( \pm 5,0)$.

Solution: Length of major axis is $2a$.

$$\begin{equation} \begin{aligned} 2a = 26 \\ a = 13 \\\end{aligned} \end{equation} $$

and coordinates of focus is $( \pm ae,0)$.

$$\begin{equation} \begin{aligned} ae = 5 \\ e = \frac{5}{{13}} \\\end{aligned} \end{equation} $$

Now, we know that $${e^2} = \ {1 - \frac{{{b^2}}}{{{a^2}}}} $$$$\begin{equation} \begin{aligned} \frac{{25}}{{169}} = 1 - \frac{{{b^2}}}{{169}} \\ {b^2} = 144 \\\end{aligned} \end{equation} $$

Therefore, the equation of ellipse in standard form is $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$$$\frac{{{x^2}}}{{169}} + \frac{{{y^2}}}{{144}} = 1$$

Question 4. Find the eccentricity of ellipse shown in figure $12$ if $S$ and $S'$ are foci and $SP$ is perpendicular to $S'P$.

Solution: Let us assume the slope of line joining points $S$ and $P$ be ${m_1}$ and the slope of line joining points $S'$ and $P$ be ${m_2}$.

Therefore, the equation of line $SP$ is $$\begin{equation} \begin{aligned} y - 0 = {m_1}(x + ae) \\ y = {m_1}(x + ae) \\\end{aligned} \end{equation} $$

and the equation of line $S'P$ is $$\begin{equation} \begin{aligned} y - 0 = {m_2}(x + ae) \\ y = {m_2}(x + ae) \\\end{aligned} \end{equation} $$

It is given that $SP$ is perpendicular to $S'P$, therefore, $${m_1} \times {m_2} = - 1$$

and $(0,b)$ is the intersection point of $SP$ and $S'P$, it satisfies the equations of line.

$$b = {m_1}ae...(1)$$ and $$b = {-m_2}ae...(2)$$

Multiply equations $(1)$ and $(2)$, we get $$\begin{equation} \begin{aligned} {b^2} = {a^2}{e^2}( - {m_1}{m_2}) \\ {b^2} = {a^2}{e^2}(1) \\ {e^2} = \frac{{{b^2}}}{{{a^2}}} \\\end{aligned} \end{equation} $$

and for ellipse, we know that $${e^2} = 1 - \frac{{{b^2}}}{{{a^2}}}$$

Therefore, $$\begin{equation} \begin{aligned} \frac{{{b^2}}}{{{a^2}}} = 1 - \frac{{{b^2}}}{{{a^2}}} \\ 2\frac{{{b^2}}}{{{a^2}}} = 1 \\ 2{e^2} = 1 \\ e = \frac{1}{{\sqrt 2 }} \\\end{aligned} \end{equation} $$