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 equation to the hyperbola drawn through the point of the standard ellipse, whose eccentric angle is $α,$ and which is confocal with the given ellipse, is $\frac{x^{2}}{\cos^{2} α} - \frac{y^{2}}{\sin^{2} α} = a^{2} - b^{2}$

Accepted Answer

To find the equation of hyperbola which is confocal with the ellipse

Equation of ellipse in standard form

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a > b)$

Here, foci $(\pm a e, 0)$ or $(\pm \sqrt{a^{2} - b^{2}}, 0)$

$(∵ e = \sqrt{1 - \frac{b^{2}}{a^{2}}})$

Let equation to any conic having the same center and axes as the given conic.

$\Rightarrow \frac{x^{2}}{A} + \frac{y^{2}}{B} = 1 (A > B)$

Here, foci $(\pm \sqrt{A - B}, 0)$

These foci are the same so $a^{2} - b^{2} = A - B$

$\Rightarrow A - a^{2} = B - b^{2} = k$ (say)

$\Rightarrow A = a^{2} + k, B = b^{2} + k$

Now the equation of the confocal conic $\frac{x^{2}}{a^{2} + k} + \frac{y^{2}}{b^{2} + k} = 1$

Ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a > b) . . . (i)$

Having an eccentric angle $α$ .

So, parameter $x = a \cos α, y = b \sin α$

Now the confocal conic is

$\frac{x^{2}}{a^{2} + k} + \frac{y^{2}}{b^{2} + k} = 1 . . . (ii)$ ( $k$ is constant)

According to the question, this confocal conic also passes through the point $(a \cos α, b \sin α)$ .

Using equation $(ii)$

$\Rightarrow \frac{a^{2} \cos^{2} α}{a^{2} + k} + \frac{b^{2} \sin^{2} α}{b^{2} + k} = 1$

$\Rightarrow a^{2} \cos^{2} α (b^{2} + k) + b^{2} \sin^{2} α (a^{2} + k) = (a^{2} + k) (b^{2} + k)$

$\Rightarrow a^{2} b^{2} \cos^{2} α + a^{2} k \cos^{2} α + a^{2} b^{2} \sin^{2} α + k b^{2} \sin^{2} α = a^{2} b^{2} + k (a^{2} + b^{2}) + k^{2}$

$\Rightarrow a^{2} b^{2} + k (a^{2} \cos^{2} α + b^{2} \sin^{2} α) = a^{2} b^{2} + k (a^{2} + b^{2}) + k^{2}$

$\Rightarrow k^{2} + k (a^{2} + b^{2} - a^{2} \cos^{2} α - b^{2} \sin^{2} α) = 0$

$\Rightarrow k (k + a^{2} (1 - \cos^{2} α) + b^{2} (1 - \sin^{2} α)) = 0$

$\Rightarrow k (k + a^{2} \sin^{2} α + b^{2} \cos^{2} α) = 0$

Either $k = 0$ or $k = - (a^{2} \sin^{2} α + b^{2} \cos^{2} α)$

Case-I: If $k = 0$

Using equation $(ii)$ , we get,

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ (Ellipse itself)

Case-II: If $k = - (a^{2} \sin^{2} α + b^{2} \cos^{2} α)$

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

$\Rightarrow \frac{x^{2}}{a^{2} - a^{2} \sin^{2} α - b^{2} \cos^{2} α} + \frac{y^{2}}{b^{2} - a^{2} \sin^{2} α - b^{2} \cos^{2} α} = 1$

$\Rightarrow \frac{x^{2}}{a^{2} (1 - \sin^{2} α) - b^{2} \cos^{2} α} + \frac{y^{2}}{b^{2} (1 - \cos^{2} α) - a^{2} \sin^{2} α} = 1$

$\Rightarrow \frac{x^{2}}{a^{2} \cos^{2} α - b^{2} \cos^{2} α} + \frac{y^{2}}{b^{2} \sin^{2} α - a^{2} \sin^{2} α} = 1$

$\Rightarrow \frac{x^{2}}{\cos^{2} α (a^{2} - b^{2})} - \frac{y^{2}}{\sin^{2} α (a^{2} - b^{2})} = 1$

$\Rightarrow \frac{x^{2}}{\cos^{2} α} - \frac{y^{2}}{\sin^{2} α} = a^{2} - b^{2}$

It is an equation of a hyperbola.

Prove that the equation to the hyperbola drawn through the point of the standard ellipse, whose eccentric angle is $α,$ and which is confocal with the given ellipse, is $\frac{x^{2}}{\cos^{2} α} - \frac{y^{2}}{\sin^{2} α} = a^{2} - b^{2}$

Important Questions on Miscellaneous Propositions

Learn and Prepare for any exam you want !

Prove that the equation to the hyperbola drawn through the point of the standard ellipse, whose eccentric angle is α, and which is confocal with the given ellipse, is x2cos2α-y2sin2α=a2-b2

Important Questions on Miscellaneous Propositions

Learn and Prepare for any exam you want !

Prove that the equation to the hyperbola drawn through the point of the standard ellipse, whose eccentric angle is $α,$ and which is confocal with the given ellipse, is $\frac{x^{2}}{\cos^{2} α} - \frac{y^{2}}{\sin^{2} α} = a^{2} - b^{2}$