AP EAMCET · Maths · Parabola
Let the equation of the tangent at a point \(\mathrm{P}\) on the parabola \(x^2-4 x-4 y+16=0\) be \(2 x-y-5=0\). If the equation of the normal drawn at \(\mathrm{P}\) to this parabola is \(\mathrm{ax}+\mathrm{y}+\mathrm{c}=0\), then \(\mathrm{ac}=\).0
- A -20
- B 20
- C 5
- D -5
Answer & Solution
Correct Answer
(D) -5
Step-by-step Solution
Detailed explanation
\(\because x^2-4 x-4 y+16=0\) \[ \Rightarrow 2 x-4-4 \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{4-2 x}{-4}=-1+\frac{1}{2} x \] \(2 x-y-5=0\) is tangent on the parabola.…
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