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NCERT Exemplar · Q46

Q.A point moves so that square of its distance from the point (3,−2)(3,-2) is numerically equal to its distance from the line 5x−12y=35x-12y=3. The equation of its locus is ____.

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Setting the square of the distance from (3,−2)(3,-2) equal to the distance from the line 5x−12y−3=05x-12y-3=0, and resolving the modulus, gives the locus 13x2+13y2−83x+64y+172=013x^2+13y^2-83x+64y+172=0.

Step 1: Translate the condition into algebra

Let P(x,y)P(x,y) be the moving point.

Square of the distance from (3,−2)(3,-2):

d12=(x−3)2+(y+2)2d_1^2=(x-3)^2+(y+2)^2

Distance from the line 5x−12y−3=05x-12y-3=0:

d2=∣5x−12y−3∣52+(−12)2=∣5x−12y−3∣13d_2=\frac{|5x-12y-3|}{\sqrt{5^2+(-12)^2}}=\frac{|5x-12y-3|}{13}

The condition given is d12=d2d_1^2=d_2.

Step 2: Set up the equation

(x−3)2+(y+2)2=∣5x−12y−3∣13(x-3)^2+(y+2)^2=\frac{|5x-12y-3|}{13}

Multiply both sides by 13 and expand the left side: …

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