एक 12 सेंटीमीटर लम्बी छड़ इस प्रकार चलती है कि इसके सिरे निर्देशांछो को स्पर्श करते हैं बिंदु का बिंदुपथ ज्ञात कीजिए जो X-अक्ष के संपर्क वाले सिरे से 3 सेमी दूर है

Options

A. `x^(2)/9+y^(2)/81=1`
B. `x^(2)/81+y^(2)/9=1`
C. `x^(2)/9+y^(2)/81=1`
D. `x^(2)/81+y^(2)/9=1`

Answer

B

Explanation

let ab be the rod of length 12 cm. let oa= a and ob =b using pythagoras theorem in `Deltaoab, a^(2)+b^(2)=(12)^(2)rArra^(2)+b^(2)=144`
` Let `AP =3cm,"then"Pm=Ab-Ap=12-3=9cm`
`therefore " " AP:PB=3:9=1:3` Let P=(h,k) for which locus to be found. if p(h,k) divides the points A and B internally in the ratio m:n, then `h=(mx_(2)+nx_1)/(m+n) and k=(my_2+ ny_1)/(m+n) `therefore h=(1x0+3xa)/(1+3)rArrh=(3a)/(4)`
`rArra=(4h)/(4)rArrb=4k`
` put the value of a =4h/3 and b= 4k in eq. (i), we get `rArr (h^(2))/(9)+k^(2)=9rArr(h^(2))/(81)+k^(2)/(9)=1` hence, the locus of a points P (h,k) is `(x^(2))/(81)+(y^(2))/(9)=1.