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The point, at shortest distance from the line  x + y = 7and lying on the ellipse  x^{2} + 2y^{2} = 6,  has coordinates 

 

Option: 1

(\sqrt{2}, \sqrt{2} )


Option: 2

(0, \sqrt{3})


Option: 3

(2,1)


Option: 4

\left(\sqrt{5}, \frac{1}{\sqrt{2}}\right)


Answers (1)

best_answer

the tangent at the point  of shortest distance

from the line $\mathrm{x}+\mathrm{y}=7 is parallel to the given line.
Any point on the given ellipse is
(\sqrt{6} \cos \theta, \sqrt{3} \sin \theta)
Equation of the tangent is


\frac{\mathrm{x} \cos \theta}{\sqrt{6}}+\frac{\mathrm{y} \sin \theta}{\sqrt{3}}=1 \text {. }
It is parallel to  x+y=7
\Rightarrow \frac{\cos \theta}{\sqrt{6}}=\frac{\sin \theta}{\sqrt{3}} \Rightarrow \frac{\cos \theta}{\sqrt{2}}=\frac{\sin \theta}{1}=\frac{1}{\sqrt{3}}
The required point is (2,1).

Posted by

Rishabh

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