TS EAMCET · Maths · Parabola
If the line \(x-y=-4 K\) is a tangent to the parabola \(y^2=8 x\) at \(P\), then the perpendicular distance of normal at \(P\) from \((K, 2 K)\) is
- A \(\frac{5}{2 \sqrt{2}}\)
- B \(\frac{7}{2 \sqrt{2}}\)
- C \(\frac{9}{2 \sqrt{2}}\)
- D \(\frac{1}{2 \sqrt{2}}\)
Answer & Solution
Correct Answer
(C) \(\frac{9}{2 \sqrt{2}}\)
Step-by-step Solution
Detailed explanation
Since, \(x-y=-4 k\) or \(y=x+4 k\) is a tangent to the parabola \(y^2=8 x\), therefore \[ \begin{aligned} 4 k & =\frac{2}{1} \quad[\because \text { Here } 4 a=8 \Rightarrow a=2] \\ k \quad k & =\frac{1}{2} \end{aligned} \] Also point of contact \(p\) is…
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