First, we need to find the focus of the given parabola y² = 4x + 16. We can rewrite the equation in the standard form Y² = 4aX. By factoring out 4 from the right side, we get y² = 4(x+4). Comparing this to the standard form, we have Y=y, X=x+4, and 4a=4, which means a=1. The focus of a parabola in the form Y²=4aX is at (X=a, Y=0).

First, we need to find the focus of the given parabola y² = 4x + 16. We can rewrite the equation in the standard form Y² = 4aX. By factoring out 4 from the right side, we get y² = 4(x+4). Comparing this to the standard form, we have Y=y, X=x+4, and 4a=4, which means a=1. The focus of a parabola in the form Y²=4aX is at (X=a, Y=0).

We know the center of the circle is (h, k) = (-3, 0) and its radius is r=5. Using the standard equation of a circle (x-h)² + (y-k)² = r², we can write the equation of circle C as (x-(-3))² + (y-0)² = 5², which simplifies to (x+3)² + y² = 25.

We know the center of the circle is (h, k) = (-3, 0) and its radius is r=5. Using the standard equation of a circle (x-h)² + (y-k)² = r², we can write the equation of circle C as (x-(-3))² + (y-0)² = 5², which simplifies to (x+3)² + y² = 25.

Since the circle passes through the intersection point, we substitute the coordinates of the intersection point into the circle's equation: (x+3)² + y² = 25. This leads to a quadratic equation in. After simplifying and rearranging terms, we get 6 ² + - 7 = 0. Factoring this quadratic equation, we find (6 +7)( -1) = 0. This gives two possible values for: _1 = -(7)/(6) and _2 = 1. Since _1 < _2, these are the correct assignments.

Since the circle passes through the intersection point, we substitute the coordinates of the intersection point into the circle's equation: (x+3)² + y² = 25. This leads to a quadratic equation in. After simplifying and rearranging terms, we get 6 ² + - 7 = 0. Factoring this quadratic equation, we find (6 +7)( -1) = 0. This gives two possible values for: _1 = -(7)/(6) and _2 = 1. Since _1 < _2, these are the correct assignments.

Finally, we need to calculate the value of 12 _1 + 29 _2. Substituting the values we found for _1 = -(7)/(6) and _2 = 1, we get 12(-(7)/(6)) + 29(1) = -14 + 29 = 15.

The focus of the parabola $y^2 = 4x + 16$ is the centre of the circle $C$ of radius 5. If the values of $\lambda$ , for which $C$ passes through the point of intersection of the lines $3x - y = 0$ and $x + \lambda y = 4$ , are $\lambda_1$ and $\lambda_2$ , $\lambda_1 < \lambda_2$ , then $12\lambda_1 + 29\lambda_2$ is equal to:

Answer: 15

Step-by-step solution

Step 1: Find the focus of the parabola

First, we need to find the focus of the given parabola $y^2 = 4x + 16$ . We can rewrite the equation in the standard form $Y^2 = 4aX$ . By factoring out 4 from the right side, we get $y^2 = 4(x+4)$ . Comparing this to the standard form, we have $Y=y$ , $X=x+4$ , and $4a=4$ , which means $a=1$ . The focus of a parabola in the form $Y^2=4aX$ is at $(X=a, Y=0)$ .

Step 2: Determine the center of the circle

Substituting $X=a=1$ and $Y=0$ into our expressions for $X$ and $Y$ , we get $x+4=1$ , which gives $x=-3$ , and $y=0$ . So, the focus of the parabola is at $(-3, 0)$ . The problem states that this focus is the center of circle $C$ . Therefore, the center of circle $C$ is $(-3, 0)$ .

Step 3: Find the equation of the circle

We know the center of the circle is $(h, k) = (-3, 0)$ and its radius is $r=5$ . Using the standard equation of a circle $(x-h)^2 + (y-k)^2 = r^2$ , we can write the equation of circle $C$ as $(x-(-3))^2 + (y-0)^2 = 5^2$ , which simplifies to $(x+3)^2 + y^2 = 25$ .

Step 4: Find the intersection point of the lines

Next, we find the point of intersection of the two given lines: $3x - y = 0$ and $x + \lambda y = 4$ . From the first equation, we get $y = 3x$ . Substituting this into the second equation gives $x + \lambda(3x) = 4$ . Factoring out $x$ , we have $x(1+3\lambda) = 4$ , so $x = \frac{4}{1+3\lambda}$ . Then, $y = 3x = \frac{12}{1+3\lambda}$ . The intersection point is $\left(\frac{4}{1+3\lambda}, \frac{12}{1+3\lambda}\right)$ .

Step 5: Substitute intersection point into circle equation to find $\lambda$

Since the circle passes through the intersection point, we substitute the coordinates of the intersection point into the circle's equation: $(x+3)^2 + y^2 = 25$ . This leads to a quadratic equation in $\lambda$ . After simplifying and rearranging terms, we get $6\lambda^2 + \lambda - 7 = 0$ . Factoring this quadratic equation, we find $(6\lambda+7)(\lambda-1) = 0$ . This gives two possible values for $\lambda$ : $\lambda_1 = -\frac{7}{6}$ and $\lambda_2 = 1$ . Since $\lambda_1 < \lambda_2$ , these are the correct assignments.

Step 6: Calculate the final expression

Finally, we need to calculate the value of $12\lambda_1 + 29\lambda_2$ . Substituting the values we found for $\lambda_1 = -\frac{7}{6}$ and $\lambda_2 = 1$ , we get $12\left(-\frac{7}{6}\right) + 29(1) = -14 + 29 = 15$ .

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