Ellipse Ques 28
- A tangent to the ellipse $x^{2}+4 y^{2}=4$ meets the ellipse $x^{2}+2 y^{2}=6$ at $P$ and $Q$. Prove that the tangents at $P$ and $Q$ of the ellipse $x^{2}+2 y^{2}=6$ are at right angles.
(1997, 5M)
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Solution:
- Given, $x^{2}+4 y^{2}=4$ or $\frac{x^{2}}{4}+\frac{y^{2}}{1}=1$……(i)
Equation of any tangent to the ellipse on (i) can be written as
$ \frac{x}{2} \cos \theta+y \sin \theta=1 $……(ii)
Equation of second ellipse is
$ \Rightarrow \quad \begin{aligned} & x^{2}+2 y^{2}=6 \\ & \Rightarrow \quad \frac{x^{2}}{6}+\frac{y^{2}}{3}=1……(iii) \end{aligned} $
Suppose the tangents at $P$ and $Q$ meets at $A(h, k)$. Equation of the chord of contact of the tangents through $A(h, k)$ is
$ \frac{h x}{6}+\frac{k y}{3}=1 $……(iv)
But Eqs. (iv) and (ii) represent the same straight line, so comparing Eqs. (iv) and (ii), we get
$ \begin{aligned} \Rightarrow \quad & \frac{h / 6}{\cos \theta / 2} & =\frac{k / 3}{\sin \theta}=\frac{1}{1} \\ & h & =3 \cos \theta \text { and } k=3 \sin \theta \end{aligned} $
Therefore, coordinates of $A$ are $(3 \cos \theta, 3 \sin \theta)$.
Now, the joint equation of the tangents at $A$ is given by $T^{2}=S S _1$,
i.e. $(\frac{h x}{6}+\frac{k y}{3}-1)^2=(\frac{x^{2}}{6}+\frac{y^{2}}{3}-1) \quad (\frac{h^{2}}{6}+\frac{k^{2}}{3}-1)$…..(v)
In Eq. (v), coefficient of $x^{2}=\frac{h^{2}}{36}-\frac{1}{6} (\frac{h^{2}}{6}+\frac{k^{2}}{3}-1)$
$ =\frac{h^{2}}{36}-\frac{h^{2}}{36}-\frac{k^{2}}{18}+\frac{1}{6}=\frac{1}{6}-\frac{k^{2}}{18} $
and coefficient of $y^{2}=\frac{k^{2}}{9}-\frac{1}{3} (\frac{h^{2}}{6}+\frac{k^{2}}{3}-1)$
$ =\frac{k^{2}}{9}-\frac{h^{2}}{18}-\frac{k^{2}}{9}+\frac{1}{3}=-\frac{h^{2}}{18}+\frac{1}{3} $
Again, coefficient of $x^{2}+$ coefficient of $y^{2}$
$ \begin{aligned} & =-\frac{1}{18}\left(h^{2}+k^{2}\right)+\frac{1}{6}+\frac{1}{3} \\ & =-\frac{1}{18}\left(9 \cos ^{2} \theta+9 \sin ^{2} \theta\right)+\frac{1}{2} \\ & =-\frac{9}{18}+\frac{1}{2}=0 \end{aligned} $
which shows that two lines represent by Eq. (v) are at right angles to each other.
BETA
