JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
The focus of the parabola \(y^2=4 x+16\) 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 \(3 x-y=0\) and \(x+\lambda y=4\), are \(\lambda_1\) and \(\lambda_2, \lambda_1 \lt \lambda_2\), then \(12 \lambda_1+29 \lambda_2\) is equal to \(\qquad\)
- A 10
- B 15
- C 20
- D 25
Answer & Solution
Correct Answer
(B) 15
Step-by-step Solution
Detailed explanation
\(y^2=4(x+4)\) Equation of circle \((x+3)^2+y^2=25\) Passes through the point of intersection of two lines \(3 x-y=0\) and \(x+\lambda y=4\) \(\left(\frac{4}{3 \lambda+1}, \frac{12}{3 \lambda+1}\right) \text {, we get }\)…
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