MHT CET · Maths · Application of Derivatives
If equation of normal to the curve \(x=\sqrt{t}\), \(y=\mathrm{t}-\frac{1}{\sqrt{\mathrm{t}}}\) at \(\mathrm{t}=4\) is
- A \(8 x+2 y=23\)
- B \(34 x-8 y=40\)
- C \(8 x+6 y=37\)
- D \(8 x+34 y=135\)
Answer & Solution
Correct Answer
(D) \(8 x+34 y=135\)
Step-by-step Solution
Detailed explanation
\(x=\sqrt{\mathrm{t}}, y=\mathrm{t}-\frac{1}{\sqrt{\mathrm{t}}}\)
\(\therefore \quad y=x^2-\frac{1}{x}\), at \(\mathrm{t}=4, x=2\) and \(y=\frac{7}{2}\)
\(\therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=2 x+\frac{1}{x^2}\)
\(\therefore \quad\) Slope of the normal at \(\mathrm{t}=4\) is \(\frac{-1}{\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)_{\mathrm{t}=4}}=\frac{-4}{17}\)
\(\therefore \quad\) Required equation is \(\left(y-\frac{7}{2}\right)=\frac{-4}{17}(x-2)\) i.e., \(8 x+34 y=135\)
\(\therefore \quad y=x^2-\frac{1}{x}\), at \(\mathrm{t}=4, x=2\) and \(y=\frac{7}{2}\)
\(\therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=2 x+\frac{1}{x^2}\)
\(\therefore \quad\) Slope of the normal at \(\mathrm{t}=4\) is \(\frac{-1}{\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)_{\mathrm{t}=4}}=\frac{-4}{17}\)
\(\therefore \quad\) Required equation is \(\left(y-\frac{7}{2}\right)=\frac{-4}{17}(x-2)\) i.e., \(8 x+34 y=135\)
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