Question

If the maximum value of $(x+y{)}^2\text{is}\lambda$ and $P(x,y)$ satisfies $x^2+y^2=1$, then the number of tangents that can drawn from $(\lambda ,0)$ to the hyperbola $(x-2{)}^2-y^2=1$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is C $2$

Any point on the circle $P(x,y)\equiv (\cos \theta ,\sin \theta )$
$\Rightarrow x=\cos \theta ,y=\sin \theta (x+y{)}^2=1+2\sin \theta \cos \theta \Rightarrow (x+y{)}^2=1+\text{sin}2\theta$
Thus maximum value of $(x+y{)}^2=\lambda =2$
Now
$S_1=(2-2{)}^2-(0{)}^2-1=-1\Rightarrow S_1<0$
So given point is external to hyperbola.
Thus number of tangents drawn from $(2,0)$ is $2$.