TS EAMCET · Maths · Parabola
If \(x-2 y+k=0\) is a tangent to the parabola \(\mathrm{y}^2-4 \mathrm{x}-4 \mathrm{y}+8=0\), then the value of \(\mathrm{K}\) is
- A \(2\)
- B \(\frac{2}{5}\)
- C \(7\)
- D \(-7\)
Answer & Solution
Correct Answer
(C) \(7\)
Step-by-step Solution
Detailed explanation
\(y=\mathrm{m} x+\mathrm{c} \Rightarrow x-2 y+k=0\) \[ \begin{aligned} & m=\frac{1}{2}, c=\frac{k}{2} \\ & y^2-4 x-4 y+8=0 \\ & (y-2)^2=4(x-1) \end{aligned} \] From (i), \(x=2 y-k\) \[ \begin{aligned} & (y-2)^2=4(2 y-k-1) \\ & y^2-12 y+(8-4 k)=0 \end{aligned} \] \(\because\)…
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