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

Answer 1

Answer: (A)

Let $A\left(x_1,x_1-5\right)$ be a point on $x-y=5$, then the chord of contact of $x^2+4y^2=4$ with respect to $\bar{A}$ is
$x\cdot x_1+4y\left(x_1-5\right)=4$
$\Rightarrow (x+4y)x_1-(20y+4)=0$
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 $\left(\frac{4}{5},-\frac{1}{5}\right)$.