Name: Pair of Tangent, Chord of Contact
Uploaded: 2019-07-19
Duration: 19 min 37 s
Description: This video contains all the details of the chord of the contact pair of tangent and cord bisected at a point. This concept is very useful for the students.

Question

Prove that the perpendicular from the focus upon any tangent on the ellipse and the line joining the center to the point of contact, meet on the corresponding directrix.

Accepted Answer

let the equation of the ellipse be $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
Let $P (a cosθ, b sinθ)$ be the point on the ellipse

Then, the equation of the tangent at the point $P (a \cos θ, b \sin θ)$ is given as,

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

$\Rightarrow \frac{x \cos θ}{a} + \frac{y \sin θ}{b} = 1 \dots (i)$

The equation of the line $S Q$ through the focus $S (a e, 0)$ and the perpendicular to equation $(i)$ can be given as,

$\frac{x \sin θ}{b} - \frac{y \cos θ}{a} - \frac{a e \sin θ}{b} = 0 . . . (i i)$

Also, the equation of $C P$ is given as,

$\frac{x \sin θ}{a} - \frac{y cosθ}{b} = 0 . . . (i i i)$ (where $C$ is the center $(0, 0)$ )

Substituting the value of $y$ from equation $(iii)$ and then, solving equations $(ii)$ and $(iii)$ , we get,

$\frac{x \sin θ}{b} - (\frac{x b \sin θ}{a \cos θ}) \frac{\cos θ}{a} - \frac{a e \sin θ}{b} = 0$

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

$\Rightarrow x e^{2} = a e$ (let $\frac{a^{2} - b^{2}}{a^{2}} = e^{2}$ )

$x = \frac{a}{e}$ (which is the equation of the directrix)

Hence, $S Q$ and $C P$ meet on the directrix.

Prove that the perpendicular from the focus upon any tangent on the ellipse and the line joining the center to the point of contact, meet on the corresponding directrix.

Important Questions on The Ellipse

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