The focus of the parabola y2=4x+16 is the centre of the circle C of radius 5. If the values of λ, for which C passes through the point of intersection of the lines 3x−y=0 and x+λy=4, are λ1 and λ2, λ1<λ2, then 12λ1+29λ2 is equal to:
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Step-by-Step Solution
Step 1: Find the focus of the parabola
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).
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=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.
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+λy=4. From the first equation, we get y=3x. Substituting this into the second equation gives x+λ(3x)=4. Factoring out x, we have x(1+3λ)=4, so x=1+3λ4. Then, y=3x=1+3λ12. The intersection point is (1+3λ4,1+3λ12).
Step 5: Substitute intersection point into circle equation to find λ
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 λ. After simplifying and rearranging terms, we get 6λ2+λ−7=0. Factoring this quadratic equation, we find (6λ+7)(λ−1)=0. This gives two possible values for λ: λ1=−67 and λ2=1. Since λ1<λ2, these are the correct assignments.
Step 6: Calculate the final expression
Finally, we need to calculate the value of 12λ1+29λ2. Substituting the values we found for λ1=−67 and λ2=1, we get 12(−67)+29(1)=−14+29=15.