Question

The locus of the mid point of the intersect of the variable line x cos a+ y sin a = p ( p is a constant) between the co-ordinate axes is

• A. $\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{p^2}$
• B. $\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{p^2}$
• C. $\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{4p^2}$
• D. None of these

Answer 1

The correct option is D

None of these

Let the locus or the mid point of intercepts be P(h,k). We will replace (h,K) with (x,y) after we get a relation between h and k. We will first find the intercepts then their mid point

(1) Finding x-intercept
y=0
$\Rightarrow x=\frac{p}{cosa}$
$\Rightarrow (\frac{P}{cosa},0)$ is the x-intercept

(2) Finding y-intercept
x=0
$\Rightarrow y=\frac{P}{sina}$
$\Rightarrow (0,\frac{P}{sina})isthey-intercept$

(3) Finding the mid-point
$P(h,k)=(\frac{\frac{p}{cosa}+0}{2},\frac{0+\frac{p}{sina}}{2})$
$=(\frac{P}{2cosa},\frac{P}{2sina})$
(4) Eliminating a (or cos a and sin a)
We will use the identity $co{s}^{2}a+si{n}^{2}a=1$ to eliminate a.
cos a=$\frac{P}{2h}$ and sin a=$\frac{P}{2k}$
$\Rightarrow \frac{P^2}{4h^2}+\frac{P^2}{4k^2}=1$
$\Rightarrow \frac{1}{h^2}+\frac{1}{k^2}=\frac{4}{p^2}$

We will replace (h,k) with (x,y)
$\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}$
$\Rightarrow$ None of the given options are correct