The distance of a point $P(h,k)$ from a pair of lines passing through the origin is $d$ units. Prove that the equation of the pair of lines is $(xk-hy)^2=d^2(x^2+y^2)$.
Please help me. I am having trouble in starting.
Help much appreciated
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Is the distance of $P$ from both lines same and equal to $d$? – Sep 06 '16 at 06:23
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It might be. I am confused too. – pi-π Sep 06 '16 at 06:26
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I did not uderstand. Please elaborate. – pi-π Sep 06 '16 at 07:03
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It would be great if you posted a solution. – pi-π Sep 06 '16 at 07:28
2 Answers
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Let equation of any of the lines be $y-mx=0$. From the given condition (using distance formula) $\dfrac{|k-mh|}{\sqrt{1+m^2}}=d$. Since this is quadratic in $m$ its roots will give the two slopes of required lines. All we have to do is to eliminate $m$ from this. $m=y/x$. Thus it becomes$(k-mh)^2=d^2(1+m^2)$, substituting $m$ we get $(k-\frac yx h)^2=d^2(1+{(\frac yx )}^2)\Rightarrow (kx-hy)^2=d^2(x^2+y^2)$.
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Consider the circle with center $(h,k)$ and radius $d$. The equation is \begin{align*} (x-h)^2 + (y-k)^2 - d^2 &= 0\\ x^2+y^2-2hx - 2ky + h^2 +k^2 - d^2 &= 0 \end{align*} We want the pair of tangents to this circle from the origin. The general form of the pair of tangents from $(x_1,y_1)$ to a second degree curve $S=0$ is $T^2 = SS_1$ where $T$ is the tangent expression for the curve. In this case, \begin{align*} T &= xx_1 + yy_1 -h(x+x_1) - k(y+y_1) + h^2 + k^2-d^2\\ S &= x^2+y^2-2hx - 2ky + h^2 +k^2 - d^2\\ S_1&= h^2+k^2-d^2 \end{align*} where $(x_1,y_1) = (0,0)$. Thus the required equation is \begin{align*} (hx +ky -(h^2+k^2-d^2))^2 = (x^2+y^2-2hx - 2ky + h^2 +k^2 - d^2)(h^2+k^2-d^2) \end{align*} and this simplifies to \begin{align*} (kx-hy)^2 &= d^2(x^2+y^2) \end{align*}
Another way to solve the problem:
The circle with center $(h,k)$ and radius $d$ is
\begin{align*}
x^2 + y^2 - 2hx - 2ky + h^2 + k^2 - d^2 = 0
\end{align*}
The circle with $(0,0)$ and $(h,k)$ as ends of diameter is
\begin{align*}
x(x-h)+y(y-k) = 0
\end{align*}
The common chord of these circles is
\begin{align*}
hx + ky = h^2+k^2-d^2
\end{align*}
If this common chord meets the circles at $A,B$, then the lines $OA, OB$ are tangents to the first circle and hence they are at a distance $d$ from the point $(h,k)$. We want to find the pair of lines joining the origin to the points of $$
x^2 + y^2 - 2(hx+ky)\left(\frac{hx+ky}{h^2 + k^2 - d^2}\right) + (h^2+k^2-d^2)\left(\frac{hx+ky}{h^2 + k^2 - d^2}\right)^2 = 0
$$
This simplifies to
$$
(kx-hy)^2 = d^2(x^2+y^2)
$$
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What is meant by $S=0$?
Is the equation of the curve and circle same?
Plz reply fast.
– pi-π Sep 06 '16 at 08:14 -
$S$ can be any second degree equation. In this case we have used the equation of the circle. – Sep 06 '16 at 08:16
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Chat feature is not supported by my cell phone and, I could not reach to a computer right now?
It would be a favour for me, if you could explain how did you wrote the equation for $T$ and $S$?
– pi-π Sep 06 '16 at 08:27 -
The expression $T=0$ is the tangent line at $(x_1,y_1)$ for the curve. This is obtained by changing $x^2$ to $xx_1$, $y^2$ to $yy_1$, $x$ to $\frac{x+x_1}{2}$, $y$ to $\frac{y+y_1}{2}$, $xy$ to $\frac{xy_1+x_1y}{2}$ and leaving the constant as is. For example, tangent to $y^2 - 4ax=0$ at $(x_1,y_1)$ will be $yy_1-2a(x+x_1) = 0$. The expression $S$ is simply the equation of the curve. I hope this is clear. – Sep 06 '16 at 08:36
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