InterviewSolution
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InterviewSolution
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A point `P`moves on a plane `x/a+y/b+z/c=1.`A plane through `P`and perpendicular to `O P`meets the coordinate axes at `A , Ba n d Cdot`If the planes through `A ,Ba n dC`parallel to the planes `x=0,y=0a n dz=0,`respectively, intersect at `Q ,`find the locus of `Qdot` |
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Answer» Correct Answer - `(1)/(ax)+ (1)/(by)+(1)/(cz)= (1)/(x^(2))+(1)/(y^(2))+ (1)/(z^(2))` The given plane is `(x)/(a)+(y)/(b)+(z)/(c)=1" "(i)` Let P(h,k,l) be the point on the plane. Then `(h)/(a)+(k)/(b)+(l)/(c)=1" "(ii)` `impliesOPsqrt(h^(2)+k^(2)+l^(2))` Direction cosines of OP are `(h)/sqrt(h^(2)+k^(2)+l^(2)),(k)/sqrt(h^(2)+k^(2)+l^(2))` and `(h)/sqrt(h^(2)+k^(2)+l^(2))` The equation of the plane through P and normal to OP is `(hx)/sqrt(h^(2)+k^(2)+l^(2))+(ky)/sqrt(h^(2)+k^(2)+l^(2))+(lz)/sqrt(h^(2)+k^(2)+l^(2))` `=sqrt(h^(2)+k^(2)+l^(2))` or `hx+ky+lz=h^(2)+k^(2)+l^(2)` Therefore, `A-=((h^(2)+k^(2)+l^(2))/(h),0,0)`, `B-=(0,(h^(2)+k^(2)+l^(2))/(k),0)` and `C-=(0,0,(h^(2)+k^(2)+l^(2))/(l))` If Q `(alpha,beta,gamma)`, then `alpha=(h^(2)+k^(2)+l^(2))/(h),beta=(h^(2)+k^(2)+l^(2))/(k)` and `gamma=(h^(2)+k^(2)+l^(2))/(l)" "(iii)` Now, `(1)/(a^(2))+(1)/(beta^(2))+(1)/(gamma^(2))` `=(h^(2)+k^(2)+l^(2))/((h^(2)+k^(2)+l^(2))^(2))=(1)/(h^(2)+k^(2)+l^(2))" "(iv)` From (iii), `h=(h^(2)+k^(2)+l^(2))/(alpha)or(h)/(a)=(h^(2)+k^(2)+l^(2))/(aalpha)` Similarly, `(k)/(b)=(h^(2)+k^(2)+l^(2))/(b beta)and(1)/(c)=(h^(2)+k^(2)+l^(2))/(cgamma)` `(h^(2)+k^(2)+l^(2))/(aalpha)+(h^(2)+k^(2)+l^(1))/(b beta)+(h^(2)+k^(2)+l^(2))/(cgamma)` `=(h)/(a)+(k)/(b)+(l)/(c)=1" "["from"(ii)]` or `(1)/(aalpha)+(1)/(b beta)+(1)/(cgamma)=(1)/(h^(2)+k^(2)+l^(2))=(1)/(alpha)+(1)/(beta^(2))+(1)/(gamma^(2))` [from (iv)] The required equation of locus is `(1)/(ax)+(1)/(by)+(1)/(cz)+(1)/(x^(2))+(1)/(y^(2))+(1)/(z^(2)` |
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