JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
If one end of a focal chord \(\mathrm{AB}\) of the parabola \(\mathrm{y}^{2}=8 \mathrm{x}\) is at \(\mathrm{A}\left(\frac{1}{2},-2\right),\) then the equation of the tangent to it at \(\mathrm{B}\) is
- A \(2 x+y-24=0\)
- B \(x-2 y+8=0\)
- C \(2 x-y-24=0\)
- D \(x+2 y+8=0\)
Answer & Solution
Correct Answer
(B) \(x-2 y+8=0\)
Step-by-step Solution
Detailed explanation
\(\mathrm{y}^{2}=8 \mathrm{x}\) \(4 \mathrm{t}_{1}=-2 \Rightarrow \mathrm{t}_{1}=-\frac{1}{2}\) \(\mathrm{t}_{1}, \mathrm{t}_{2}=-1\) \(\mathrm{t}_{2}=-\frac{1}{\mathrm{t}_{1}}\) \(\Rightarrow \mathrm{t}_{2}=2\) So coordinate of \(\mathrm{B}\) is \((8,8)\) \(\therefore\)…
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