JEE Advanced · Mathematics · 13. Parabola
Let \(L\) be a normal to the parabola \(y^2=4 x\). If \(L\) passes through the point \((9,6)\), then \(L\) is given by
- A
\(y-x+3=0\)
- B
\(y+3 x-33=0\)
- C
\(y+x-15=0\)
- D
\(y-2 x+12=0\)
Answer & Solution
Correct Answer
(A)
\(y-x+3=0\)
Step-by-step Solution
Detailed explanation
Normal to \(y^2=4 x\), is \(y-m x-2 m-m^3\) which passes through \((9,6)\).
Now, \(\quad 6=9 m-2 m-m^3\) \(\Rightarrow m^3-7 m+6=0 \Rightarrow m=1,2,-3\)
\(\therefore\) Equation of normals are
\(y-x+3=0\)
\(y+3 x-33=0\) and \(y-2 x+12=0\)
Now, \(\quad 6=9 m-2 m-m^3\) \(\Rightarrow m^3-7 m+6=0 \Rightarrow m=1,2,-3\)
\(\therefore\) Equation of normals are
\(y-x+3=0\)
\(y+3 x-33=0\) and \(y-2 x+12=0\)
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