TS EAMCET · Maths · Parabola
If the line \(2 x+3 y+n=0\) is a tangent to the parabola \(y^2=8 x\), then the equation of the normal drawn at the point \((2 n, 4 \sqrt{n})\) to the parabola \(y^2=8 x\) is
- A \(x-3 y+18=0\)
- B \(3 \mathrm{x}+2 \mathrm{y}-30=0\)
- C \(3 x+y-66=0\)
- D \(2 x-3 y+6=0\)
Answer & Solution
Correct Answer
(C) \(3 x+y-66=0\)
Step-by-step Solution
Detailed explanation
Equation of tangent to \(y^2=4 a x\) is \(y=m x+\frac{a}{m}\) \[ \therefore y=m x+\frac{2}{m} \] Comparing with \(2 x+3 y+n=0\)…
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