Let A be a point on the x-axis.

Question

Let $A$ be a point on the $x$-axis. Common tangents are drawn from $A$ to the curves $x^2+y^2=8$ and $y^2=16 x$. If one of these tangents touches the two curves at $Q$ and $R$, then $(Q R)^2$ is equal to :

Accepted Answer

Let a tangent on ${y^2} = 16x$ be $$y = mx + {4 \over m}$$

For common to ${x^2} + {y^2} = 8$

$${4 \over m} = 2\sqrt 2 (1 + {m^2})$$

$$ \Rightarrow {2 \over {{m^2}}} = 1 + {m^2} \Rightarrow m = \, \pm 1$$

Taking one of the tangent $y = x + 4$

Point of tangency with ${y^2} = 4x$

$${x^2} + 8 + 16 = 4x \Rightarrow x = 4$$ \& $y = 8$

$$\therefore$$ $Q(4,8)$

and for ${x^2} + {y^2} = 8$

$2{x^2} + 8x + 8 = 0$

$${x^2} + 4x + 4 = 8 \Rightarrow x =  - 2,y = 2 \Rightarrow R = ( - 2,2)$$

${(QR)^2} = {6^2} + {6^2} = 72$

Solutions · 1