AP EAMCET · Maths · Parabola
Let a focal chord \(12 x+5 y-27=0\) of the parabola \(\mathrm{y}^2=\mathrm{kx}\) intersect the parabola at the points \(\mathrm{P}\) and \(\mathrm{P}^{\prime}\). If \(\mathrm{S}\) is the focus of this parabola, then \(9\left(\mathrm{SP}+\mathrm{SP}^1\right)=\)
- A \(27\)
- B \(108\)
- C 16 SP.SP \(^{\prime}\)
- D \(4 \mathrm{SP}. \mathrm{SP}^{\prime}\)
Answer & Solution
Correct Answer
(D) \(4 \mathrm{SP}. \mathrm{SP}^{\prime}\)
Step-by-step Solution
Detailed explanation
Given that equation of parabola \(y^2=\mathrm{km}\) then focus is \(\left(\frac{k}{4}, 0\right)\) which is lies on focal chord. \(\therefore 12 \times \frac{k}{4}+0-27=0 \Rightarrow k=9\), focus s \(\left(\frac{9}{4}, 0\right)\) On solving the equation \(y^2=9 x\) and…
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