Ellipse Ques 16
- The locus of the foot of perpendicular drawn from the centre of the ellipse $x^{2}+3 y^{2}=6$ on any tangent to it is
(a) $\left(x^{2}-y^{2}\right)^{2}=6 x^{2}+2 y^{2}$
(2014 Main)
(b) $\left(x^{2}-y^{2}\right)^{2}=6 x^{2}-2 y^{2}$
(c) $\left(x^{2}+y^{2}\right)^{2}=6 x^{2}+2 y^{2}$
(d) $\left(x^{2}+y^{2}\right)^{2}=6 x^{2}-2 y^{2}$
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Answer:
Correct Answer: 16.(c)
Solution:
- Equation of ellipse is $x^{2}+3 y^{2}=6$ or $\frac{x^{2}}{6}+\frac{y^{2}}{2}=1$.
Equation of the tangent is $\frac{x \cos \theta}{a}+\frac{y \sin \theta}{b}=1$
Let $(h, k)$ be any point on the locus.
$\therefore \quad \frac{h}{a} \cos \theta+\frac{k}{b} \sin \theta=1$
Slope of the tangent line is $\frac{-b}{a} \cot \theta$.
Slope of perpendicular drawn from centre $(0,0)$ to $(h, k)$ is $k / h$.
Since, both the lines are perpendicular.
$ \begin{array}{rlrl} \therefore & & (\frac{k}{h}) \times (-\frac{b}{a} \cot \theta ) & =-1 \\ \Rightarrow & \frac{\cos \theta}{h a} & =\frac{\sin \theta}{k b}=\alpha \\ \Rightarrow & & \cos \theta & =\alpha h a \\ \sin \theta & =\alpha k b \end{array} $
From Eq. (i), $\frac{h}{a}(\alpha h a)+\frac{k}{b}(\alpha k b)=1$
$ \begin{array}{rlrl} \Rightarrow & & h^{2} \alpha+k^{2} \alpha & =1 \\ \Rightarrow & \alpha & =\frac{1}{h^{2}+k^{2}} \end{array} $
Also, $\quad \sin ^{2} \theta+\cos ^{2} \theta=1$
$\Rightarrow \quad(\alpha k b)^{2}+(\alpha h a)^{2}=1$
$\Rightarrow \quad \alpha^{2} k^{2} b^{2}+\alpha^{2} h^{2} a^{2}=1$
$\Rightarrow \frac{k^{2} b^{2}}{\left(h^{2}+k^{2}\right)^{2}}+\frac{h^{2} a^{2}}{\left(h^{2}+k^{2}\right)^{2}}=1$
$\Rightarrow \quad \frac{2 k^{2}}{\left(h^{2}+k^{2}\right)^{2}}+\frac{6 h^{2}}{\left(h^{2}+k^{2}\right)^{2}}=1 \quad\left[\because a^{2}=6, b^{2}=2\right]$
$\Rightarrow \quad 6 x^{2}+2 y^{2}=\left(x^{2}+y^{2}\right)^{2}$
[replacing $k$ by $y$ and $h$ by $x$]
BETA
