TS EAMCET · Maths · Parabola
If the normal at one end of the latus rectum of the parabola \(y^2=16 x\) meets the \(X\)-axis at the point \(P\), then the length of the chord passing through \(P\) and perpendicular to the normal is
- A \(48 \sqrt{2}\)
- B \(32 \sqrt{2}\)
- C \(24 \sqrt{2}\)
- D \(20 \sqrt{2}\)
Answer & Solution
Correct Answer
(B) \(32 \sqrt{2}\)
Step-by-step Solution
Detailed explanation
We have, \(y^2=16 x\) Coordinate of latusrectum \((4,8)\) Equation of normal at \((4,8)\) is \(y=-x+12\) It cuts \({x}\)-axis at \((12,0)\) Equation of chord passing through \((12,0)\) and perpendicular of normal is \(y=x-12\) Put the value of \(y\) in \(y^2=16 x\), we get…
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