MHT CET · Maths · Parabola
If \(2 x+y+\lambda=0\) is normal to the parabola \(y^{2}=8 x\), then \(\lambda\) is
- A \(-24\)
- B 8
- C \(-16\)
- D 24
Answer & Solution
Correct Answer
(A) \(-24\)
Step-by-step Solution
Detailed explanation
Given, parabola, \(y^{2}=8 x\)
\(\Rightarrow 2 y \frac{d y}{d x}=8\)
\(\Rightarrow \frac{d y}{d x}=\frac{4}{y}\)
Slope of normal of parabola \(=-\frac{y}{4}\) \(\ldots(\mathrm{i})\)
Given, \(2 x+y+\lambda=0\) is a equation of normal of the parabola \(y^{2}=8 x\).
\(\therefore\) Slope of normal \(=-2\) \(\ldots\) (ii)
From Eqs. (i) and (ii), we get
\(\therefore\) \(-\frac{y}{4}=-2 \Rightarrow y=8\)
\((8)^{2}=8 x \Rightarrow x=8\)
Now, putting the values of \(x\) and \(y\) in the equation of normal
\(2(8)+8+\lambda =0\)
\(\Rightarrow \lambda =-24\)
\(\Rightarrow 2 y \frac{d y}{d x}=8\)
\(\Rightarrow \frac{d y}{d x}=\frac{4}{y}\)
Slope of normal of parabola \(=-\frac{y}{4}\) \(\ldots(\mathrm{i})\)
Given, \(2 x+y+\lambda=0\) is a equation of normal of the parabola \(y^{2}=8 x\).
\(\therefore\) Slope of normal \(=-2\) \(\ldots\) (ii)
From Eqs. (i) and (ii), we get
\(\therefore\) \(-\frac{y}{4}=-2 \Rightarrow y=8\)
\((8)^{2}=8 x \Rightarrow x=8\)
Now, putting the values of \(x\) and \(y\) in the equation of normal
\(2(8)+8+\lambda =0\)
\(\Rightarrow \lambda =-24\)
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