Question

Tangent are drawn from the points on the line $x-y-5=0$ to $x^2+4y^2=4$, then all the chords of contact pass through a fixed point, whose coordinate are

• A. $(\frac{4}{5},-\frac{1}{5})$
• B. $(\frac{4}{5},\frac{1}{5})$
• C. $(-\frac{4}{5},\frac{1}{5})$
• D. None of these

Answer 1

The correct option is A $(\frac{4}{5},-\frac{1}{5})$

Let $A(x_1,x_1-5)$ be a point of $x-y=5$, then the chord of contact of $x^2+4y^2=4$ with respect to $A$ is

$x\cdot x_1+4y(x_1-5)=4$

$\Rightarrow (x+4y)x_1-(20y+4)=0$

Since it passes through a fixed point.

$\therefore x+4y=0$

and $20y+4=0$ (from $P+\lambda Q=0)$

$\Rightarrow y=-\frac{1}{5}$ and $x=\frac{4}{5}$

So, the coordinates of fixed point is $(\frac{4}{5},-\frac{1}{5})$.