Question

prove that the locus of the midpoint of the portion of line x cos alpha + y Sin Alpha equals to p which is intercepted between the axes given that Alpha is variable is -- one upon X square + 1 upon y square equals to 4 upon p square​

Answer 1

Answer:

$\bf{\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}}$

Step-by-step explanation:

Let P(h,k) be the midpoint of the portion of the line

$x\:cos\alpha+y\:sin\alpha=p$.....(1)

put y=0 in (1), we get

$x\:cos\alpha=p$

$x=\frac{p}{cos\alpha}$

put x=0 in(1), we get

$y\:sin\alpha=p$

$y=\frac{p}{sin\alpha}$

Therefore, the line (1) meets the coordinate axes at

A($\frac{p}{cos\alpha},0$)  and  B(0,$\frac{p}{sin\alpha}$)

Clearly, the midpoint of the Portion AB= P

$(\frac{\frac{p}{cos\alpha}+0}{2},\frac{0+\frac{p}{sin\alpha}}{2})=(h,k)$

$(\frac{p}{2cos\alpha},\frac{p}{2sin\alpha})=(h,k)$

$\implies\:h=\frac{p}{2cos\alpha},\:\:\:\:k=\frac{p}{2sinalpha}$

$\implies\:cos\alpha=\frac{p}{2h},\:\:\:\:sin\alpha=\frac{p}{2k}$

squaring and adding these equations, we get

$cos^2\alpha+sin^2\alpha=\frac{p^2}{4h^2}+\frac{p^2}{4k^2}$

$\implies\:1=\frac{p^2}{4h^2}+\frac{p^2}{4k^2}$

$\implies\:1=\frac{p^2}{4}(\frac{1}{h^2}+\frac{1}{k^2})$

$\implies\:\frac{4}{p^2}=\frac{1}{h^2}+\frac{1}{k^2}$

$\therefore\text{The locus of P is}$

$\boxed{\bf{\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}}}$

Answer 2

Answer:

Step-by-step explanation: