Name: Parametric Equation of an Ellipse
Uploaded: 2019-07-19
Duration: 7 min 12 s
Description: In this video, you will understand the parametric equation of an ellipse with examples, which will help you solve questions in the exam. Watch and learn

Question

Prove that the length of the focal chord of the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

which is inclined to the major axis at angle

θ

is

\frac{2 a b^{2}}{a^{2} \sin^{2} θ + b^{2} \cos^{2} θ}

.

Accepted Answer

Given that, the equation of the ellipse is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

Let $P S Q$ be the focal chord of the ellipse

Therefore, $S = (a e, 0)$ and $S^{'} = (- a e, 0)$ , where $e$ is eccentricity of the ellipse

Now, the slope of the focal chord $= \tan θ$

So, the equation of $P S Q$ is $y - 0 = \tan θ (x - a e)$

$\Rightarrow y = \tan θ (x - a e)$

Let $P (x_{1} y_{1}) and Q (x_{2}, y_{2})$ are points on the ellipse

Therefore, $\frac{x^{2}}{a^{2}} + \frac{\tan^{2} θ (x - a e)^{2}}{b^{2}} = 1$

$\Rightarrow b^{2} x^{2} + a^{2} \tan^{2} θ (x - a e)^{2} = a^{2} b^{2}$

$\Rightarrow b^{2} x^{2} + a^{2} \tan^{2} θ (x^{2} - 2 a e x + a^{2} e^{2}) - a^{2} b^{2} = 0$

$\Rightarrow b^{2} x^{2} + a^{2} \tan^{2} θ x^{2} - 2 a^{3} e \tan^{2} θ x + a^{4} e^{2} \tan^{2} θ - a^{2} b^{2} = 0$

$\Rightarrow (b^{2} + a^{2} \tan^{2} θ) x^{2} - (2 a^{3} e \tan^{2} θ) x + a^{4} e^{2} \tan^{2} θ - a^{2} b^{2} = 0 \to (1)$

It is a quadratic equation in $x$ , therefore, sum of roots $x_{1} + x_{2} = \frac{2 a^{3} e \tan^{2} θ}{b^{2} + a^{2} \tan^{2} θ}$ and product of roots $x_{1} x_{2} = \frac{a^{4} e^{2} \tan θ - a^{2} b^{2}}{b^{2} + a^{2} \tan θ}$ [Assume that, $x_{1}, x_{2}$ are the roots of equation $(1)$ ]

Now, $y = \tan θ [x - a e]$

So, $y_{1} = \tan θ [x_{1} - a e]$ and $y_{2} = \tan θ [x_{2} - a e]$

Thus, $y_{1} - y_{2} = \tan θ [x_{1} - a e] - \tan θ [x_{2} - a e]$

$= \tan θ [x_{1} - a e - x_{2} + a e]$

$= \tan θ [x_{1} - x_{2}]$

Therefore, the length of focal chord $P Q = \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}}$

$= \sqrt{{(x_{1} - x_{2})}^{2} + {[\tan θ (x_{1} - x_{2})]}^{2}}$

$= \sqrt{(x_{1} - x_{2})^{2} + \tan^{2} θ {(x_{1} - x_{2})}^{2}}$

$= \sqrt{{(x_{1} - x_{2})}^{2} [1 + \tan^{2} θ]}$

$= \sqrt{{(x_{1} - x_{2})}^{2} \cdot (\sec^{2} θ)}$

$= (x_{1} - x_{2}) (\sec θ)$

$= \sqrt{{(x_{1} + x_{2})}^{2} - 2 x_{1} x_{2}} \cdot (\sec θ)$

$= \sqrt{{(\frac{2 a^{3} e \tan^{2} θ}{b^{2} + a^{2} \tan^{2} θ})}^{2} - 4 \times \frac{a^{4} e^{2} \tan θ - a^{2} b^{2}}{b^{2} + a^{2} \tan θ}} \cdot \sec θ$

$= \sqrt{\frac{4 a^{6} e^{2} \tan^{4} θ}{{(b^{2} + a^{2} + \tan^{2} θ)}^{2}} - \frac{4 a^{4} e^{2} \tan^{2} θ - 4 a^{2} b^{2}}{b^{2} + a^{2} \tan^{2} θ}} \cdot \sec θ$

$= \sqrt{\frac{4 a^{6} e^{2} \tan^{4} θ - (4 a^{4} e^{2} \tan^{2} θ - 4 a^{2} b^{2}) (b^{2} + a^{2} \tan^{2} θ)}{{(b^{2} + a^{2} \tan^{2} θ)}^{2}}} \cdot \sec θ$

$= \sqrt{4 a^{6} e^{2} \tan^{4} θ - (4 a^{4} b^{2} e^{2} \tan^{2} θ + 4 a^{6} e^{2} \tan^{4} θ - 4 a^{2} b^{4} - 4 a^{4} b^{2} + 4 a^{4} b^{2} \tan^{2} θ)} \cdot \frac{\sec θ}{b^{2} + a^{2} \tan^{2} θ}$

$= \sqrt{4 a^{2} b^{4} + 4 a^{4} b^{2} \tan^{2} θ - 4 a^{4} b^{2} e^{2} \tan^{2} θ} \cdot \frac{\sec θ}{b^{2} + a^{2} \tan^{2} θ}$

$= \sqrt{4 a^{2} b^{2} [b^{2} + a^{2} \tan^{2} θ - a^{2} e^{2} \tan^{2} θ]} \cdot \frac{\sec θ}{b^{2} + a^{2} \tan^{2} θ}$

$= \sqrt{b^{2} + a^{2} \tan^{2} θ - a^{2} e^{2} \tan^{2} θ} \cdot \frac{2 a b \sec θ}{b^{2} + a^{2} \tan^{2} θ}$

$= \sqrt{b^{2} + a^{2} \tan^{2} θ - a^{2} [\frac{a^{2} - b^{2}}{a^{2}}] \tan^{2} θ} \cdot \frac{2 a b \sec θ}{b^{2} + a^{2} \tan^{2} θ}$ [Using $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{\frac{a^{2} - b^{2}}{a^{2}}}$ ]

$= \sqrt{b^{2} + a^{2} \tan^{2} θ - a^{2} \tan^{2} θ + b^{2} \tan^{2} θ} \cdot \frac{2 a b secθ}{b^{2} + a^{2} \tan θ}$

$= \sqrt{b^{2} (\sec^{2} θ)} \cdot \frac{2 a b \sec θ}{b^{2} + a^{2} \tan^{2} θ}$

$= b \sec θ \cdot \frac{2 a b \sec θ}{b^{2} + a^{2} \tan^{2} θ}$

$= \frac{2 a b^{2} \sec^{2} θ}{b^{2} + a^{2} \tan^{2} θ}$

Now, dividing numerator and denominator with $\sec^{2} θ$

$= \frac{2 a b^{2}}{a^{2} \sin^{2} θ + b^{2} \cos^{2} θ}$

Hence, proved.

Prove that the length of the focal chord of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ which is inclined to the major axis at angle $θ$ is $\frac{2 a b^{2}}{a^{2} \sin^{2} θ + b^{2} \cos^{2} θ}$ .

Important Questions on Ellipse

Learn and Prepare for any exam you want !

Prove that the length of the focal chord of the ellipse x2a2+y2b2=1 which is inclined to the major axis at angle θ is 2ab2a2sin2θ+b2cos2θ.

Important Questions on Ellipse

Learn and Prepare for any exam you want !

Prove that the length of the focal chord of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ which is inclined to the major axis at angle $θ$ is $\frac{2 a b^{2}}{a^{2} \sin^{2} θ + b^{2} \cos^{2} θ}$ .