Ellipse 2 Question 10

10. 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: 10. (c)

Solution:

  1. 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, $\quad \sin ^{2} \theta+\cos ^{2} \theta=1$

$\Rightarrow \quad(\alpha k b)^{2}+(\alpha h a)^{2}=1$

$\Rightarrow \quad \alpha^{2} k^{2} b^{2}+\alpha^{2} h^{2} a^{2}=1$

$\Rightarrow \frac{k^{2} b^{2}}{\left(h^{2}+k^{2}\right)^{2}}+\frac{h^{2} a^{2}}{\left(h^{2}+k^{2}\right)^{2}}=1$

$\Rightarrow \quad \frac{2 k^{2}}{\left(h^{2}+k^{2}\right)^{2}}+\frac{6 h^{2}}{\left(h^{2}+k^{2}\right)^{2}}=1 \quad\left[\because a^{2}=6, b^{2}=2\right]$

$\Rightarrow \quad 6 x^{2}+2 y^{2}=\left(x^{2}+y^{2}\right)^{2}$



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