Question

The equation of the locus of a point which is at a distance of 5 units from a fixed point (1,4) and also from a fixed line 2x+3y-1=0 is

• A. $9x^2+12xy+4y^2-30x-108y+222=0$
• B. $9x^2-12xy+4y^2-30x-98y+220=0$
• C. $9x^2+12xy+4y^2-22x-108y+222=0$
• D. $9x^2-12xy+4y^2-22x-98y+220=0$

Hint:

When a locus problem describes the distance from a point being equal to the distance from a line, immediately identify it as a parabola. The formula $PS^2 = PM^2$ (where PM is the perpendicular distance to the directrix) is the most direct way to find its equation. Be careful with the expansion of the squared terms.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The question asks for the locus of a point whose distance from a fixed point is equal to its distance from a fixed line. This is the definition of a parabola. The fixed point is the focus and the fixed line is the directrix. The phrase "at a distance of 5 units" appears to be extraneous information or a mistake in the problem statement, as it contradicts the standard definition of a locus which should be a curve, not discrete points. We will proceed by finding the equation of the parabola.
Step 2: Key Formula or Approach
Let the point on the locus be $P(x,y)$. Let the focus be $S(1,4)$. Let the directrix be the line $L: 2x+3y-1=0$. The definition of a parabola states that the distance $PS$ is equal to the perpendicular distance from $P$ to the line $L$. \[ PS = \text{Distance}(P, L) \] Squaring both sides gives $PS^2 = (\text{Distance}(P, L))^2$. Distance formula: $PS^2 = (x-x_1)^2 + (y-y_1)^2$. Perpendicular distance from a point $(x,y)$ to a line $Ax+By+C=0$ is $\frac{|Ax+By+C|}{\sqrt{A^2+B^2}}$.
Step 3: Detailed Explanation
Using the formula with $P(x,y)$, focus $S(1,4)$, and directrix $2x+3y-1=0$: \[ \sqrt{(x-1)^2 + (y-4)^2} = \frac{|2x+3y-1|}{\sqrt{2^2+3^2}} \] Square both sides of the equation: \[ (x-1)^2 + (y-4)^2 = \frac{(2x+3y-1)^2}{4+9} \] \[ (x^2-2x+1) + (y^2-8y+16) = \frac{(2x+3y-1)^2}{13} \] Multiply both sides by 13: \[ 13(x^2+y^2-2x-8y+17) = (2x+3y-1)^2 \] Expand both sides: \[ 13x^2+13y^2-26x-104y+221 = (2x)^2+(3y)^2+(-1)^2+2(2x)(3y)+2(3y)(-1)+2(-1)(2x) \] \[ 13x^2+13y^2-26x-104y+221 = 4x^2+9y^2+1+12xy-6y-4x \] Now, move all terms to one side to get the final equation: \[ (13x^2-4x^2) - 12xy + (13y^2-9y^2) + (-26x+4x) + (-104y+6y) + (221-1) = 0 \] \[ 9x^2 - 12xy + 4y^2 - 22x - 98y + 220 = 0 \] Step 4: Final Answer
The equation of the locus (parabola) is $9x^2-12xy+4y^2-22x-98y+220=0$. This matches option (D).