JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let the focal chord \(P Q\) of the parabola \(y^2=4 x\) make an angle of \(60^{\circ}\) with the positive \(x\)-axis, where P lies in the first quadrant. If the circle, whose one diameter is PS, S being the focus of the parabola, touches the \(y\)-axis at the point \((0, \alpha)\), then \(5 \alpha^2\) is equal to :
- A 15
- B 25
- C 30
- D 20
Answer & Solution
Correct Answer
(A) 15
Step-by-step Solution
Detailed explanation
\(\tan 60^{\circ}=\frac{2 \mathrm{t}-0}{\mathrm{t}^2-1}=\sqrt{3} \Rightarrow \mathrm{t}=\sqrt{3}\) \(\therefore \mathrm{P}(3,2 \sqrt{3})\) Circle : \((x-1)(x-3)+(y-0)(y-2 \sqrt{3})=0\) at \(x=0\) \(\Rightarrow 3+y^2-2 \sqrt{3} y=0\) \(\Rightarrow \mathrm{y}=\sqrt{3}=\alpha\)…
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