Question

From any point on the line $y=x+4$, tangents are drawn to the auxiliary circle of the ellipse $x^2+4y^2=4.$ If $P,Q$ are the points of contact and $A,B$ are the corresponding points of $P$ and $Q$ on the ellipse respectively, then the locus of the midpoint of $AB$ is

• A. $4x^2+y^2+y-2x=0$
• B. $4x^2+y^2+y+2x=0$
• C. $4y^2+x^2+x-2y=0$
• D. $4y^2+x^2+x+2y=0$

Answer 1

The correct option is C $4y^2+x^2+x-2y=0$

Given equation of ellipse is
$\frac{x^2}{4}+\frac{y^2}{1}=1\dots \left(1\right)$
auxiliary circle is $x^2+y^2=4\dots \left(2\right)$
Let $P\equiv (x_1,y_1),Q\equiv (x_2,y_2),A\equiv (x_1^{\prime },y_1^{\prime }),B\equiv (x_2^{\prime },y_2^{\prime })$
Now, $x_1=x_1^{\prime }$ and $x_2=x_2^{\prime }$
By equation $\left(1\right)$ and $\left(2\right)$
and $y_1=2y_1^{\prime },y_2=2y_2^{\prime }$
and $M(h,k)$ be the midpoint of $AB$.
any point on the line $y=x+4$ will be $R(t,t+4)$

Equation of chord $PQ$ is
$tx+(t+4)y=4\Rightarrow x=\frac{4-(t+4)y}{t}$
from equation $\left(2\right)$
$16+(t+4{)}^2{y}^2-8(t+4)y+t^2{y}^2=4t^2\Rightarrow [t^2+(t+4{)}^2]y^2-8(t+4)y+16-4t^2=0$
$y_1+y_2=\frac{8(t+4)}{t^2+(t+4{)}^2}$
similarly by replacing $y$ we get
$x_1+x_2=\frac{8t}{t^2+(t+4{)}^2}$
Now $h=\frac{x_1+x_2}{2}=\frac{4t}{t^2+(t+4{)}^2}$
$k=\frac{y_1^{\prime }+y_2^{\prime }}{2}=\frac{y_1+y_2}{4}=\frac{2(t+4)}{t^2+(t+4{)}^2}$
$\therefore \frac{h^2}{4}+k^2=\frac{4[t^2+(t+4{)}^2]}{[t^2+(t+4{)}^2{]}^2}=\frac{4}{t^2+(t+4{)}^2}$
and $\frac{k}{2}-\frac{h}{4}=\frac{4}{t^2+(t+4{)}^2}$
Hence locus will be
$\frac{x^2}{4}+y^2=\frac{y}{2}-\frac{x}{4}$
$4y^2+x^2+x-2y=0.$