Question

The equation of the curve obtained by reflecting the ellipse $\frac{(x-4)^2}{16} + \frac{(y-3)^2}{9} = 1$ about the line $x - y - 2 = 0$ is $16x^2 + 9y^2 + k_1x - 36y + k_2 = 0$ then sum of prime factors of $(k_1 + k_2)$ is:

Answer 1

16

Answer 2

1. Reflection Transformation:

   For a line $x - y - 2 = 0$ (with $a=1$, $b=-1$, $c=-2$), the reflection of a point $(X,Y)$ to $(x,y)$ is given by:

   $$x = Y + 2,\quad y = X - 2.$$

   Inverting, we have:

   $$X = y + 2,\quad Y = x - 2.$$

2. Substitute in the Ellipse Equation:

   Original ellipse:

   $$\frac{(X-4)^2}{16} + \frac{(Y-3)^2}{9} = 1.$$

   Substitute $X = y+2$ and $Y = x-2$:

   $$\frac{((y+2)-4)^2}{16} + \frac{((x-2)-3)^2}{9} = 1 \quad\Longrightarrow\quad \frac{(y-2)^2}{16} + \frac{(x-5)^2}{9} = 1.$$

3. Bring to Standard Form:

   Multiply through by 144 (LCM of 16 and 9):

   $$9(y-2)^2 + 16(x-5)^2 = 144.$$

   Expanding:

   $$9(y^2 - 4y + 4) + 16(x^2 - 10x + 25) = 144,$$ $$9y^2 - 36y + 36 + 16x^2 - 160x + 400 = 144.$$

   Combine like terms:

   $$16x^2 + 9y^2 - 160x - 36y + (36+400-144) = 0,$$ $$16x^2 + 9y^2 - 160x - 36y + 292 = 0.$$

4. Identification:

   The equation is given in the form:

   $$16x^2 + 9y^2 + k_1x - 36y + k_2 = 0.$$

   Comparing:

   $$k_1 = -160 \quad\text{and}\quad k_2 = 292.$$

   Then:

   $$k_1 + k_2 = -160 + 292 = 132.$$

5. Sum of Prime Factors of 132:

   Factorize:

   $$132 = 2^2 \times 3 \times 11.$$

   Sum of distinct prime factors:

   $$2 + 3 + 11 = 16.$$

Explanation of the Solution:

Reflect the ellipse via the formula to get new coordinates. Substitute $X = y+2$ and $Y = x-2$ in the ellipse equation, simplify and match coefficients to find $k_1$ and $k_2$. Then calculate $k_1+k_2=132$ and sum its distinct prime factors $2, 3, 11$ to get $16$.