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  • The line cuts the ellipse <img alt="\mathrm{ x^2+4 y^2=1}" src="https:/

The line \mathrm{x+2 y=1} cuts the ellipse \mathrm{ x^2+4 y^2=1} at two distinct points A and B. Point C is on the ellipse such that area of triangle ABC is maximum then find point C.

Option: 1

\mathrm{\left(-\frac{1}{\sqrt{2}},-\frac{1}{2 \sqrt{2}}\right)}


Option: 2

\mathrm{\left(-\frac{1}{2}, 1\right)}


Option: 3

\mathrm{\left(\frac{1}{\sqrt{2}}, \frac{1}{2 \sqrt{2}}\right)}


Option: 4

\mathrm{\left(-\frac{1}{2},-\frac{1}{2}\right)}


Answers (1)

best_answer

  1. Clearly, line \mathrm{x+2 y=1} cuts the ellipse \mathrm{\frac{x^2}{1}+\frac{y^2}{1 / 4}} at \mathrm{A(1,0)} and \mathrm{B\left(0, \frac{1}{2}\right).}

Let point C on the ellipse be \mathrm{\left(\cos \theta, \frac{\sin \theta}{2}\right).}

Now area of triangle ABC is maximum, if distance of C from AB is maximum.

Distance of C from AB,

\mathrm{ d=\left|\frac{\cos \theta+\sin \theta-1}{\sqrt{5}}\right| }

Maximum value of d occurs when \mathrm{\cos \theta=\sin \theta=-\frac{1}{\sqrt{2}}.}

\mathrm{ \therefore \quad C \equiv\left(-\frac{1}{\sqrt{2}},-\frac{1}{2 \sqrt{2}}\right) }

Posted by

Shailly goel

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