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Tangents are drawn from the points on the line  \mathrm x-\mathrm y-5=0 \text { to } \mathrm x^2+4 \mathrm y^2=4Then all the chords of contact pass through a fixed point, whose coordinates are

Option: 1

\left(\frac{4}{5},-\frac{1}{5}\right)


Option: 2

\left(\frac{1}{5},-\frac{4}{5}\right)


Option: 3

\left(\frac{4}{5}, \frac{1}{5}\right)


Option: 4

\left(-\frac{4}{5}, \frac{1}{5}\right)


Answers (1)

best_answer

Any point on the line \mathrm{x}-\mathrm{y}-5=0 will be of the form (\mathrm{t}, \mathrm{t}-5).  Chord of contact of this point w.r.t curve   \mathrm{x}^2+4 \mathrm{y}^2=4   is   \mathrm{tx}+4(\mathrm{t}-5) \mathrm{y}-4=0    or (-20\mathrm{y}-4)+\mathrm{t}(\mathrm{x}+4 \mathrm{y})=0  which is a family of straight lines, each member of this family pass through the point of intersection of straight lines -20 \mathrm y-4=0   and   \mathrm x+ 4 \mathrm{y}=0. 

Hence, (a) is the correct answer.

Posted by

vishal kumar

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