Question

15. A line remains at distance of 5 units from origin and intersects the axes in points 'x^(2)+y^(2)=25' and '_|_' then(A) locus mid-point of '(1)/(x^(2))+(1)/(y^(2))=(4)/(25)' is 'PQ' (B) locus foot of 'Q' from origin to the line is 'P' (C) the line makes with both the axes an area which cannot exceed 25 square units(D) none of these​

Answer 1

Answer:

Equation of plane at a distance 1 from origin is given by,

lx+my+nz=1, where l,m,n are direction cosine of normal to the plane along distance

Thus intercept on the axes are,

A=(

l

1

,0,0),B=(0,

m

1

,0),C=(0,0,

n

1

)

So centroid of ABC is,

x=

3l

1

,y=

3m

1

,z=

3n

1

Also we know,

l

2

+m

2

+n

2

=1

⇒

9x

2

1

+

9y

2

1

+

9z

2

1

=1

⇒

x

2

1+y 21 + z21=9

Hence, required value of k is 9.

Step-by-step explanation:

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