A rod of length eight units moves such that its ends $A$ and $B$ always lie on the lines $x-y+2=0$ and $y+2=0$, respectively. If the locus of the point $P$, that divides the rod $AB$ internally in the ratio $2:1$ is $9\left(x^2 + \alpha y^2 + \beta xy + \gamma x + 28 y\right) - 76 = 0$, then $\alpha - \beta - \gamma$ is equal to:

Step 1: Define coordinates of A, B, and P

Let the coordinates of point $A$ be $(x_A, y_A)$ and point $B$ be $(x_B, y_B)$. Since $A$ lies on the line $x-y+2=0$, we have $y_A = x_A+2$. Since $B$ lies on the line $y+2=0$, we have $y_B = -2$. Let $P(x, y)$ be the point that divides $AB$ in the ratio $2:1$.

Step 2: Apply Section Formula

Using the section formula for internal division, we can express $x_A, y_A, x_B, y_B$ in terms of $x, y$. From the equations, we get $x_A = 3x - 2x_B$ and $y_A = 3y - 2y_B$.

Step 3: Substitute coordinates into line equations

Substitute $y_A = x_A+2$ and $y_B = -2$ into the expressions from the section formula. This gives us $3y - 2(-2) = (3x - 2x_B) + 2$, which simplifies to $3y + 4 = 3x - 2x_B + 2$. Rearranging this, we find $2x_B = 3x - 3y - 2$, so $x_B = \frac{3x - 3y - 2}{2}$.

Step 4: Calculate $x_A$ and $y_A$

Now we can find $x_A$ and $y_A$ using the expressions derived earlier. Substituting $x_B$ into the equation for $x_A$, we get $x_A = 3x - (3x - 3y - 2) = 3y + 2$. Then, using $y_A = x_A+2$, we find $y_A = (3y+2)+2 = 3y+4$.

Step 5: Use distance formula for AB

The length of the rod $AB$ is 8 units. We use the distance formula to set up an equation: $(x_A - x_B)^2 + (y_A - y_B)^2 = 8^2$. Substitute the expressions for $x_A, y_A, x_B, y_B$ in terms of $x$ and $y$ into this equation.

Step 6: Substitute and simplify to find locus

Substitute the expressions for $x_A, y_A, x_B, y_B$ into the distance formula. This gives us: $$\left( (3y+2) - \frac{3x-3y-2}{2} \right)^2 + \left( (3y+4) - (-2) \right)^2 = 64$$ Simplify the terms inside the parentheses: $$\left( \frac{6y+4-3x+3y+2}{2} \right)^2 + (3y+6)^2 = 64$$ $$\left( \frac{-3x+9y+6}{2} \right)^2 + (3y+6)^2 = 64$$ $$\frac{9(-x+3y+2)^2}{4} + 9(y+2)^2 = 64$$ Multiply by 4: $$9(-x+3y+2)^2 + 36(y+2)^2 = 256$$ Expand the squares: $$9(x^2 + 9y^2 + 4 - 6xy - 4x + 12y) + 36(y^2 + 4y + 4) = 256$$ $$9x^2 + 81y^2 + 36 - 54xy - 36x + 108y + 36y^2 + 144y + 144 = 256$$ Combine like terms: $$9x^2 + 117y^2 - 54xy - 36x + 252y + 180 = 256$$ $$9x^2 + 117y^2 - 54xy - 36x + 252y - 76 = 0$$ Divide by 9: $$x^2 + 13y^2 - 6xy - 4x + 28y - \frac{76}{9} = 0$$ Multiply by 9 again to match the given form: $$9(x^2 + 13y^2 - 6xy - 4x + 28y) - 76 = 0$$ Comparing this with the given locus $9(x^2 + \alpha y^2 + \beta xy + \gamma x + 28 y) - 76 = 0$, we find: $\alpha = 13$, $\beta = -6$, $\gamma = -4$.

Step 7: Calculate $\alpha - \beta - \gamma$

Now we substitute the values of $\alpha = 13$, $\beta = -6$, and $\gamma = -4$ into the expression $\alpha - \beta - \gamma$.

More from this chapter

• Find the equation of the circle passing through points (1,2), (3,4), and (5,2).
• Determine the equation of the circle with radius 4 and Centre (-2, 3).
• Find the equation of the hyperbola where foci are (0, +-6) and the length of the latus rectum is 36.
• Find the equation of the hyperbola where foci are (0, ± 12) and the length of the latus rectum is 36
• Find the equation of the circle with centre (-3,2) and radius 10.
• Find the equation of the circle with centre (-3,2) and radius 9.
• Find the equation of the circle with centre (-3,2) and radius 8.
• Find the equation of the circle with centre (-3,2) and radius 6.

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