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 circle on any focal distance of the ellipse as diameter touches the auxiliary circle.

Accepted Answer

Let any point $P$ be $(a \cos θ, b \sin θ)$ on the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a > b)$
and the focus $S^{'}$ be $(- a e, 0)$ as shown in the figure.

The equation to the circle drawn with $S^{'} P$ as diameter
$(x - a \cos θ) (x + a e) + (y - b s i n θ) (y - 0) = 0$
$\Rightarrow S_{1} \equiv x^{2} + y^{2} + x (a e - a \cos θ) - b \sin θ y - a^{2} e \cos θ = 0 . . . (i)$

and equation of the auxiliary circle is

$S_{2} \equiv x^{2} + y^{2} = a^{2} . . . (ii)$

Equation of radical axis is $S_{1} - S_{2} = 0$ $\Rightarrow x (a e - a \cos θ) - b \sin θ y + a^{2} - a^{2} e \cos θ = 0 . . . (iii)$

We have to prove that the radical axis must touch the auxiliary circle $(ii)$ at $Q$ as shown in the figure.
We apply the condition of tangency.
Length of perpendicular from $(0,0)$ on $(iii)$
$d = |\frac{a^{2} - a^{2} e \cos θ}{\sqrt{[a^{2} {(e - \cos θ)}^{2} + b^{2} \sin^{2} θ]}}|$
$\Rightarrow d = |\frac{a^{2} (1 - e \cos θ)}{\sqrt{a^{2} (e^{2} + \cos^{2} θ - 2 e \cos θ) + a^{2} (1 - e^{2}) \sin^{2} θ}}|$
$\Rightarrow d = |\frac{a (1 - e \cos θ)}{\sqrt{(e^{2} (1 - \sin^{2} θ) + (\cos^{2} θ + \sin^{2} θ) - 2 e \cos θ)}}|$
$\Rightarrow d = |\frac{a (1 - e \cos θ)}{\sqrt{(e^{2} \cos^{2} θ + 1 - 2 e \cos θ)}}|$

$\Rightarrow d = |\frac{a (1 - e \cos θ)}{\sqrt{{(1 - e \cos θ)}^{2}}}|$
$\Rightarrow d = |\frac{a (1 - e \cos θ)}{(1 - e \cos θ)}|$
$\Rightarrow d = a =$ radius of the circle.
Hence, the circle $(i)$ and $(ii)$ touch each other.

Hence proved.

Prove that the circle on any focal distance of the ellipse as diameter touches the auxiliary circle.

Important Questions on The Ellipse

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