Home
Class 12
MATHS
A B C is an equilateral triangle with A(...

`A B C` is an equilateral triangle with `A(0,0)` and `B(a ,0)` , (a>0). L, M and `N` are the foot of the perpendiculars drawn from a point `P` to the side `A B ,B C ,a n dC A` , respectively. If `P` lies inside the triangle and satisfies the condition `P L^2=P MdotP N ,` then find the locus of `Pdot`

Text Solution

Verified by Experts


The equations of lines AC and BC are, respectively, given by
`y-sqrt(3) x =0`
`y+sqrt(3)(x-a) =0`
Let P(h,k) be the coordinates of the point whose locus is to be found. Now, according to the given condition, we have
`PL^(2) = PM*PN`
`i.e., k^(2) = (|k-sqrt(3)h|)/(2) * (|k+sqrt(3)(h-a)|)/(2)`
`i.e., 4k^(2) = (k-sqrt(3)h)[k+sqrt(3)(h-a)] " " [because "P lies below both the lines"]`
`i.e., 3(h^(2)+k^(2))-3ah + sqrt(3) ak = 0`
`i.e., h^(2)+k^(2)-ah + (a)/(sqrt(3)) k = 0`
Putting (x,y) in place of (h,k) gives the equation of the required locus as
` x^(2)+y^(2)-ax + (a)/(sqrt(3))y = 0`
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

M is the foot of the perpendicular from a point P on a parabola y^2=4a x to its directrix and S P M is an equilateral triangle, where S is the focus. Then find S P .

The equations of the perpendicular bisectors of the sides A Ba n dA C of triangle A B C are x-y+5=0 and x+2y=0 , respectively. If the point A is (1,-2) , then find the equation of the line B Cdot

P is a point on the straight line 7x-4y-29=0 . If the foot of the perpendicular drawn from P on the line 5x+2y+18=0 be N(-2,-4) , find the coordinates of P.

Let A B C be a triangle with A B=A Cdot If D is the midpoint of B C ,E is the foot of the perpendicular drawn from D to A C ,a n dF is the midpoint of D E , then prove that A F is perpendicular to B Edot

The coordinates of the vertices Ba n dC of a triangle A B C are (2, 0) and (8, 0), respectively. Vertex A is moving in such a way that 4tan(B/2)tan(C/2)=1. Then find the locus of A

The coordinates of the points A and B are (2,4) and (2,6) respectively . The point P is on that side of AB opposite to the origin . If PAB be an equilateral triangle , find the coordinates of P.

The coordinates of the point A and B are (a,0) and (-a ,0), respectively. If a point P moves so that P A^2-P B^2=2k^2, when k is constant, then find the equation to the locus of the point Pdot

If an a triangle A B C , b=3 c, a n d C-B=90^0, then find the value of tanB

The coordinates of the vertices A, B and C of a triangle ABC are (0,5), (-1,-2) and (11,7) respectively . Find the coordinates of the foot of the perpendicular from B on AC.