Home
Class 12
MATHS
If the line (x/a)+(y/b)=1 moves in such ...

If the line `(x/a)+(y/b)=1` moves in such a way that `(1/(a^2))+(1/(b^2))=(1/(c^2))` , where `c` is a constant, prove that the foot of the perpendicular from the origin on the straight line describes the circle `x^2+y^2=c^2dot`

Text Solution

Verified by Experts

Given variable line is `(x)/(a) + (y)/(b) = 1 " " (1)`
Line perpendicular to (1) and passing through the origin is
`(x)/(b) - (y)/(a) = 0 " " (2)`
Now, foot of perpendicular P(h,k) from the origin upon the line (1) is the point of intersection of line (1) and (2).
` (h)/(a) + (k)/(b) =1 " " (3)`
`" and " (h)/(b) - (k)/(a) =0 " " (4)`
Squaring and adding (3) and (4), we get
`h^(2) ((1)/(a^(2))+ (1)/(b^(2))) +k^(2) ((1)/(b^(2))+ (1)/(a^(2))) = 1`
`"But, it is given that " (1)/(a^(2)) + (1)/(b^(2)) = (1)/(c^(2))`
`therefore (h^(2)+ k^(2))(1)/(c^(2)) = 1`
Hence, equation of locus of P is `x^(2) + y^(2) = c^(2)`.
Promotional Banner

Topper's Solved these Questions

  • STRAIGHT LINES

    CENGAGE PUBLICATION|Exercise EXAMPLE|12 Videos

  • STRAIGHT LINES

    CENGAGE PUBLICATION|Exercise CONCEPT APPLICATION EXERCISE 2.1|23 Videos

  • STRAIGHT LINE

    CENGAGE PUBLICATION|Exercise Multiple Correct Answers Type|8 Videos

  • THEORY OF EQUATIONS

    CENGAGE PUBLICATION|Exercise JEE ADVANCED (Numerical Value Type )|1 Videos

Similar Questions

Explore conceptually related problems

The variable coefficients a, b in the equation of the straight line (x)/(a) + (y)/(b) = 1 are connected by the relation (1)/(a^(2)) + (1)/(b^(2)) = (1)/(c^(2)) where c is a fixed constant. Show that the locus of the foot of the perpendicular from the origin upon the line is a circle. Find the equation of the circle.

Find the coordinates of the foot of the perpendicular drawn from the point (2,-5) on the straight line x=y+1.

The straight line y=mx+c cuts the circle x^2+y^2=a^2 at real points if

if P is the length of perpendicular from origin to the line x/a+y/b=1 then prove that 1/(a^2)+1/(b^2)=1/(p^2)

The product of the perpendiculars from origin to the pair of lines a x^2+2h x y+b y^2 + 2gx + 2fy + c =0 is

Find the coordinates of the foot of the perpendicular drawn from the point P(1,-2) on the line y = 2x +1. Also, find the image of P in the line.

A plane passes through a fixed point (a ,b ,c)dot Show that the locus of the foot of the perpendicular to it from the origin is the sphere x^2+y^2+z^2-a x-b y-c z=0.

The perpendicular from the origin to the line y = mx + c meets it at the point (-1, 2) . Find the values of m and c.

prove that the product of the perpendiculars drawn from the point (x_1, y_1) to the pair of straight lines ax^2+2hxy+by^2=0 is |(ax_1^2+2hx_1y_1+by_1^2)/sqrt((a-b)^2+4h^2)|

If a^(x)=b^(y)=c^(z) and b^(2)=ac prove that (1)/(x)+(1)/(z)=(2)/(y)