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The locus of the foot of perpendicular drawn from the centre of the ellipse $x^{2}+3y^{2}=6$ on any tangent to it is :

Option: 1
$\left ( x^{2} +y^{2}\right )^{2}=6x^{2}+2y^{2}$

Option: 2
$\left ( x^{2} +y^{2}\right )^{2}=6x^{2}-2y^{2}$

Option: 3
$\left ( x^{2} -y^{2}\right )^{2}=6x^{2}+2y^{2}$

Option: 4
$\left ( x^{2} -y^{2}\right )^{2}=6x^{2}-2y^{2}$

Answers (1)

best_answer

Equation of ellipse is $\frac{x^{2}}{6}+\frac{y^{2}}{2}=1$

Equation of tangent is

$\frac{x\cos\theta }{a}+\frac{y\sin\theta }{b}\:=1\quad \quad \ldots(i)$

Slope = $\frac{-b}{a}\:\cot\theta$

Slope of perpendicular drawn from the centre

(0,0) to (h,k) is $\frac{k}{h}$

$\frac{k}{h}\times\:(\frac{-b}{a}\cot\theta )\:=-1$

$\frac{k}{h}=\frac{b}{a}\cot\theta$

$\frac{\cos\theta }{ah}\:=\frac{\sin\theta }{bk}$

_so,

$\frac{\cos\theta }{ah}\:=\frac{\sin\theta }{bk}\:=m\quad(say)$ (Put in (i))

$\frac{h}{a} (mah)+\frac{K}{b}(mbK)=1$

$mh^2+mK^2=1$

$m\:=\frac{1}{h^{2}+K^{2}}$

also $\sin^{2}\theta +\cos^{2}\theta \:=1$

$\sin\theta=mbk\;\;\text{and }\;\;\cos\theta=mah$

On solving, we get

$K^2b^2+h^2a^2 = (h^2+K^2)^2$

Put a² = 6, b² = 2

We get 6x²+2y² = (x²+y²)²

Posted by

Ritika Jonwal

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