MHT CET · Maths · Application of Derivatives
If \(x+y=k\) is normal to \(y^{2}=12 x\), then \(k\) is
- A 3
- B 9
- C \(-9\)
- D \(-3\)
Answer & Solution
Correct Answer
(B) 9
Step-by-step Solution
Detailed explanation
Let \(x+y=k\) be normal to
\(
y^{2}=12 x \text { at } P(\alpha, \beta)
\)
\(\therefore \quad \beta^{2}=12 \alpha\)
Also, slope of normal at \(P(\alpha, \beta)\) is \(-1\)
Form, Eq.(i),
\(\frac{d y}{d x} =\frac{6}{y} \)
\( \Rightarrow \left(\frac{d y}{d x}\right)_{P} =\frac{6}{\beta}\)
\(\therefore\) \(
-1=\frac{-1}{6 / \beta}
\)
\(\Rightarrow\) \(\beta=6, \alpha=3\)
\(\therefore P\) is \((3,6)\) which lies on \(x+y=k\)
\(\therefore 3+6=k\)
\(\Rightarrow k=9\)
\(
y^{2}=12 x \text { at } P(\alpha, \beta)
\)
\(\therefore \quad \beta^{2}=12 \alpha\)
Also, slope of normal at \(P(\alpha, \beta)\) is \(-1\)
Form, Eq.(i),
\(\frac{d y}{d x} =\frac{6}{y} \)
\( \Rightarrow \left(\frac{d y}{d x}\right)_{P} =\frac{6}{\beta}\)
\(\therefore\) \(
-1=\frac{-1}{6 / \beta}
\)
\(\Rightarrow\) \(\beta=6, \alpha=3\)
\(\therefore P\) is \((3,6)\) which lies on \(x+y=k\)
\(\therefore 3+6=k\)
\(\Rightarrow k=9\)
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