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  • The coordinates of all the points on the ellipse <img alt="\mathrm{{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%7B%5Cfrac%7Bx%5E2%7D%7Ba%5

The coordinates of all the points on the ellipse \mathrm{{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}} for which the area of the \triangle PON is the maximum where O is the origin and N is the foot of the perpendicular from O to the tangent at P, are \mathrm{\left( \pm \frac{a^2}{k}, \pm \frac{b^2}{k}\right)} where K=

Option: 1

\mathrm{a^2+b^2}


Option: 2

\mathrm{\sqrt{a^2+b^2}}-


Option: 3

\mathrm{a^2-b^2}-


Option: 4

\mathrm{\sqrt{a^2-b^2}}-


Answers (1)

best_answer


Let P be the point \mathrm{(a \cos \theta, b \sin \theta)} tangents and normals at P are \mathrm{ \frac{x \cos \theta}{a}+\frac{y \sin \theta}{b}=1 }and  \mathrm{ a x \sec \theta-b y \operatorname{cosec} \theta=a^2-b^2 \cdot } \mathrm{ \triangle \mathrm{PON}= \frac{1}{2} ON.NP =\frac{1}{2} ON. NM =\frac{1}{2} p_1 p_2 }where \mathrm{ p_1 } and \mathrm{ p_2 } are the length of perpendiculars from the origin O to tangent and normal respectively.

\begin{aligned} & \mathrm{p_1=\frac{1}{\sqrt{\frac{\cos ^2 \theta}{a^2+\frac{\sin ^2 \theta}{b^2}}} \text { and } \quad p_2=\frac{a^2-b^2}{\sqrt{\frac{a^2}{\cos ^2 \theta}+\frac{b^2}{\sin ^2 \theta}}}}} \\ & \mathrm{p_1 p_2=\frac{a^2-b^2}{\sqrt{1+1+\frac{a^2}{b^2} \tan ^2 \theta+\frac{b^2}{a^2} \cot ^2 \theta}}=\frac{a^2-b^2}{\frac{a}{b} \tan \theta+\frac{b}{a} \cot \theta}} \end{aligned}

for  \mathrm{p_1 p_2 } to be maximum denominators should be minimum.
 

\begin{aligned} &\mathrm{=\frac{a}{b} \tan \theta-\frac{b}{a} \cot \theta=0 \Rightarrow \tan ^2 \theta=\frac{b^2}{a^2} \Rightarrow \tan \theta= \pm \frac{b}{a}}\\ &\mathrm{ =\quad \sin \theta= \pm \frac{b}{\sqrt{a^2+b^2}} \text { and } \cos \theta= \pm \frac{a}{\sqrt{a^2+b^2}}} \\ & =\text { Hence the points } \mathrm{P \text { is }\left( \pm \frac{a^2}{\sqrt{a^2+b^2}}, \pm \frac{b^2}{\sqrt{a^2+b^2}}\right)} \end{aligned}

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Divya Prakash Singh

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