Question

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, then locus of P is :

• A. parabola
• B. pair of lines
• C. circle
• D. straight line

Answer 1

Answer: (A)

parabola

Answer 2

Let P be the point $(h, k)$. The equation of the circle is $x^2 + y^2 = 1$, with center at the origin $O(0,0)$ and radius $r=1$.

The equation of the tangents from $P(h, k)$ to the circle $x^2+y^2=1$ can be represented by $y = mx \pm \sqrt{1+m^2}$, where $m$ is the slope. Since the tangents pass through $P(h, k)$, we have $k = mh \pm \sqrt{1+m^2}$. Rearranging this equation yields $k - mh = \pm \sqrt{1+m^2}$. Squaring both sides gives $(k - mh)^2 = 1+m^2$, which expands to $k^2 - 2mkh + m^2h^2 = 1+m^2$. Rearranging terms to form a quadratic equation in $m$: $m^2(h^2 - 1) - 2mkh + (k^2 - 1) = 0$. The roots of this quadratic equation, $m_1$ and $m_2$, represent the slopes of the two tangents from P.

The given line is $x = 1$. Let $Q$ and $R$ be the points where the tangents $T_1$ and $T_2$ intersect the line $x=1$. The equation of a tangent line passing through $P(h, k)$ with slope $m$ is $y - k = m(x - h)$. For the intersection point $Q$ on the line $x=1$, we substitute $x=1$: $y_Q - k = m_1(1 - h)$, so $y_Q = k + m_1(1 - h)$. Thus, $Q = (1, k + m_1(1 - h))$. Similarly, for the intersection point $R$ on the line $x=1$: $y_R - k = m_2(1 - h)$, so $y_R = k + m_2(1 - h)$. Thus, $R = (1, k + m_2(1 - h))$.

The length of the intercept made by the tangents on the line $x=1$ is the distance $QR$. $QR = |y_Q - y_R| = |(k + m_1(1 - h)) - (k + m_2(1 - h))| = |(m_1 - m_2)(1 - h)|$. We are given that $QR = 2$, so $|(m_1 - m_2)(1 - h)| = 2$.

From the quadratic equation $m^2(h^2 - 1) - 2mkh + (k^2 - 1) = 0$, we can find the sum and product of the roots: Sum of roots: $m_1 + m_2 = \frac{2kh}{h^2 - 1}$ Product of roots: $m_1 m_2 = \frac{k^2 - 1}{h^2 - 1}$

We use the identity $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2$: $(m_1 - m_2)^2 = \left(\frac{2kh}{h^2 - 1}\right)^2 - 4\left(\frac{k^2 - 1}{h^2 - 1}\right)$ $(m_1 - m_2)^2 = \frac{4k^2h^2}{(h^2 - 1)^2} - \frac{4(k^2 - 1)(h^2 - 1)}{(h^2 - 1)^2}$ $(m_1 - m_2)^2 = \frac{4k^2h^2 - 4(k^2h^2 - k^2 - h^2 + 1)}{(h^2 - 1)^2}$ $(m_1 - m_2)^2 = \frac{4k^2h^2 - 4k^2h^2 + 4k^2 + 4h^2 - 4}{(h^2 - 1)^2}$ $(m_1 - m_2)^2 = \frac{4(h^2 + k^2 - 1)}{(h^2 - 1)^2}$ Taking the square root, we get $|m_1 - m_2| = \frac{2\sqrt{h^2 + k^2 - 1}}{|h^2 - 1|}$.

Now, substitute this into the intercept length equation $|(m_1 - m_2)(1 - h)| = 2$: $\left|\frac{2\sqrt{h^2 + k^2 - 1}}{|h^2 - 1|} (1 - h)\right| = 2$. $\frac{2\sqrt{h^2 + k^2 - 1}}{|h^2 - 1|} |1 - h| = 2$. Dividing by 2: $\frac{\sqrt{h^2 + k^2 - 1}}{|h^2 - 1|} |h - 1| = 1$. Since $|h^2 - 1| = |(h-1)(h+1)| = |h-1||h+1|$, we can rewrite the equation as: $\frac{\sqrt{h^2 + k^2 - 1}}{|h-1||h+1|} |h-1| = 1$.

For the tangents to be well-defined and distinct, and for the slopes to be finite, we must have $h^2 - 1 \ne 0$, which means $h \ne 1$ and $h \ne -1$. Also, for tangents to be drawn from P, P must be outside the circle, so $h^2+k^2-1 > 0$. Assuming $h \ne 1$, we can cancel $|h-1|$ from the numerator and denominator: $\frac{\sqrt{h^2 + k^2 - 1}}{|h+1|} = 1$. This implies $\sqrt{h^2 + k^2 - 1} = |h+1|$.

Squaring both sides of the equation: $h^2 + k^2 - 1 = (h+1)^2$. $h^2 + k^2 - 1 = h^2 + 2h + 1$. $k^2 - 1 = 2h + 1$. $k^2 = 2h + 2$. $k^2 = 2(h+1)$.

Replacing $(h, k)$ with $(x, y)$ to represent the locus of point P, we get: $y^2 = 2(x+1)$.

This is the standard form of a parabola. The vertex is at $(-1, 0)$ and it opens to the right. The conditions $h \ne 1$ and $h \ne -1$ are necessary for the derivation steps involving $|h^2-1|$ and for the intercept to be a well-defined segment of length 2. These conditions do not exclude any part of the derived parabolic locus itself, as the points where $h=1$ or $h=-1$ on the parabola do not satisfy the original problem's geometric constraints for a well-defined intercept of length 2. Therefore, the locus of P is a parabola.