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Two tangents are drawn from a point $P$ to the circle $x^2+y^2=1$. If the tangents make an intercept of $2$ on the line $x-1=0$ , then locus of $P$ is?
$A)$ Parabola
$B)$ Pair Of lines
$C)$ Circle
$D)$ Straight Line
What i did was to assume two tangents from point $P$ as $$y=mx \pm \sqrt{m^2+1}$$
and equate it with $x=1$. Sure it gives a value but how do i decide the ordinate then? Should I use symmetry or there is a better sophisticated method for this one?
Let $A$ and $B$ be points on $x=1$ such that $|\overline{AB}| = 2$. With $R$ the point where $x=1$ meets the circle, define $\alpha := \angle ROA$ and $\beta := \angle ROB$. Let the "other" tangent from $A$ meet the circle at $S$; we see that $\angle ROS = 2\alpha$. Likewise, the "other" tangent from $B$ meets the circle at $T$ such that $\angle ROT = 2\beta$. Let these "other" tangents intersect at $P$, and note that $\overline{OP}$ bisects $\angle SOT$. Consequently, we can write $$\begin{align} P &= \sec(\alpha-\beta)\;\left( \cos(\alpha+\beta), \sin(\alpha+\beta) \right) \\ &= \left( \frac{\cos\alpha \cos\beta - \sin\alpha\sin\beta}{\cos\alpha\cos\beta + \sin\alpha \sin\beta}, \frac{\sin\alpha \cos\beta+\cos\alpha\sin\beta}{\cos\alpha\cos\beta + \sin\alpha \sin\beta} \right) \\ &=\left( \frac{1 - \tan\alpha\tan\beta}{1 + \tan\alpha \tan\beta}, \frac{\tan\alpha+\tan\beta}{1 + \tan\alpha \tan\beta} \right) \end{align}$$
We can get the equation for the locus by eliminating $a := \tan\alpha$ and $b := \tan\beta$ from the system $$x = \frac{1-a b}{1+ab} \qquad y = \frac{a+b}{1+ab}$$ subject to the "intercept condition" $$a - b = 2$$
Without too much trouble, we arrive at the relation
$$ 2( 1 + x ) = y^2$$
which describes a parabola. $\square$
This is not a solution. The following figure is just my guess on the meaning of the question.
