JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let \(\mathrm{P}(\alpha, \beta)\) be a point on the parabola \(\mathrm{y}^2=4 \mathrm{x}\). If \(\mathrm{P}\) also lies on the chord of the parabola \(x^2=8 y\) whose mid point is \(\left(1, \frac{5}{4}\right)\). Then \((\alpha-28)(\beta-8)\) is equal to ...........
- A \(123\)
- B \(451\)
- C \(192\)
- D \(125\)
Answer & Solution
Correct Answer
(C) \(192\)
Step-by-step Solution
Detailed explanation
Parabola is \(x^2=8 y\) Chord with mid point \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) is \(\mathrm{T}=\mathrm{S}_1\) \( \therefore \mathrm{xx}_1-4\left(\mathrm{y}_1 \mathrm{y}_1\right)=\mathrm{x}_1^2-8 \mathrm{y}_1\)…
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