The longest distance of the point $(a, 0)$ from the curve $2 x^2+y^2=2 x$ is

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Aicharya · Accepted Answer

Given, curve is $2 x^2+y^2=2 x$ $2 x^2-2 x+y^2=0$ $\Rightarrow \quad 2\left(x-\frac{1}{2}ight)^2+y^2=\frac{1}{2}$ $\Rightarrow \quad \frac{\left(x-\frac{1}{2}ight)^2}{\frac{1}{4}}+\frac{y^2}{\frac{1}{2}}=1$ which represents an ellipse. Here, $a=\frac{1}{2}, b=\frac{1}{\sqrt{2}}, h=\frac{1}{2}, k=0$ Consider a point $P(h+a \cos 	heta, k+b \sin 	heta)$ $=P\left(\frac{1}{2}+\frac{1}{2} \cos 	heta, \frac{1}{\sqrt{2}} \sin 	hetaight)$ on the ellipse from which the distance of point $(a, 0)$ is maximum. let $Q(a, 0)$ Now, $P Q=\sqrt{\left(\frac{1}{2}+\frac{1}{2} \cos 	heta-aight)^2+\left(\frac{1}{\sqrt{2}} \sin 	heta-0ight)^2}$ $=\sqrt{\frac{1}{4}+\frac{1}{4} \cos ^2 	heta+a^2+\frac{1}{2} \cos 	heta-a \cos 	heta-a+\frac{1}{2} \sin ^2 	heta}$ $P Q=\sqrt{\frac{1}{2}+a^2-a+\left(\frac{1}{2}-aight) \cos 	heta+\frac{1}{4} \sin ^2 	heta}$ $\Rightarrow$ Let $y=P Q^2$ $=\frac{1}{2}+a^2-a+\left(\frac{1}{2}-aight) \cos 	heta+\frac{1}{4} \sin ^2 	heta$ For maxima and minima, put $\frac{d y}{d 	heta}=0$ $\Rightarrow-\left(\frac{1}{2}-aight) \sin 	heta+\frac{1}{4} \cdot 2 \sin 	heta \cos 	heta=0$ $\Rightarrow \quad \sin 	heta\left(-\frac{1}{2}+a+\frac{1}{2} \cos 	hetaight)=0$ $\Rightarrow \sin 	heta=0$ or $-\frac{1}{2}+a+\frac{1}{2} \cos 	heta=0$ $\Rightarrow \quad 	heta=0$ or $\cos 	heta=1-2 a$ $\Rightarrow \quad \sin ^2 	heta=1-\cos ^2 	heta$ $=1-(1-2 a)^2$ $=-4 a^2+4 a$ Now, $\frac{d^2 y}{d 	heta^2} < 0$ for $\cos 	heta=1-2 a$ Thus, distance $P Q$ is maximum, when $\cos 	heta=1-2 a$ and $\sin ^2 	heta=-4 a^2+4 a$ Now, required longest distance is $=\sqrt{\frac{1}{2}+a^2-a+\left(\frac{1}{2}-aight)(1-2 a)+\frac{1}{4}\left(-4 a^2+4 aight)}$ $=\sqrt{\frac{1}{2}+a^2-a+\frac{1}{2}-a-a+2 a^2-a^2+a}$ $=\sqrt{2 a^2-2 a+1}$ $=\sqrt{1-2 a+2 a^2}$

The longest distance of the point $(a, 0)$ from the curve $2 x^2+y^2=2 x$ is

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