Question

If the distance of a variable point \(P\) from a fixed point \((3,-4)\) is \(\dfrac23\) times the distance of \(P\) from the fixed line \(x-y+2=0\), then the locus of \(P\) is \[ ax^2+4by+by^2-62x+80y+c=0, \] then \(2c=\)

• A. \(31ab\)
• B. \(31(a+b)\)
• C. \(7ab\)
• D. \(7(a+b)\)

Hint:

A conic can be defined as the locus of a point whose distance from a fixed point (focus) bears a constant ratio to its perpendicular distance from a fixed line (directrix). Use \[ \boxed{ \text{Distance from }ax+by+c=0 = \frac{|ax+by+c|}{\sqrt{a^2+b^2}}. } \]

Answer 1

The Correct Option is B

Step 1: Use the definition of the locus. Let \[ P(x,y). \] Distance from the focus \((3,-4)\) is \[ \sqrt{(x-3)^2+(y+4)^2}. \] Distance from the directrix \[ x-y+2=0 \] is \[ \frac{|x-y+2|}{\sqrt2}. \] Given, \[ \sqrt{(x-3)^2+(y+4)^2} = \frac23 \cdot \frac{|x-y+2|}{\sqrt2}. \]

Step 2: Square both sides. Squaring, \[ (x-3)^2+(y+4)^2 = \frac29(x-y+2)^2. \] Multiplying by \(9\), \[ 9(x-3)^2+9(y+4)^2 = 2(x-y+2)^2. \] Expanding, \[ 7x^2+4xy+7y^2-62x+80y+193=0. \] Comparing with \[ ax^2+4xy+by^2-62x+80y+c=0, \] we obtain \[ a=7,\qquad b=7,\qquad c=193. \]

Step 3: Find \(2c\). Thus, \[ 2c=2(193)=386. \] Also, \[ 31(a+b) = 31(7+7) = 31\times14 = 386. \] Hence, \[ \boxed{2c=31(a+b).} \] Therefore, the correct option is \(\boxed{(B)}\).