MHT CET · Maths · Application of Derivatives
If \(y=4 x-5\) is a tangent to the curve \(y^2=\mathrm{p} x^3+\mathrm{q}\) at \((2,3)\), then the values of p and q are respectively
- A \(-2,7\)
- B \(7,-2\)
- C \(2,-7\)
- D \(-7,-2\)
Answer & Solution
Correct Answer
(C) \(2,-7\)
Step-by-step Solution
Detailed explanation
\(y^2=\mathrm{p} x^3+\mathrm{q}...(i)\)
Differentiating both sides w.r.t. \(x\), we get
\(\begin{aligned}
& 2 y \cdot \frac{\mathrm{~d} y}{\mathrm{~d} x}=3 \mathrm{p} x^2 \\
& \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{3 \mathrm{p}}{2}\left(\frac{x^2}{y}\right) \\
\therefore \quad & \left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{(2,3)}=\frac{3 \mathrm{p}}{2} \times \frac{4}{3}=2 \mathrm{p}
\end{aligned}\)
Since the line touches the curve, their slopes are equal.
\(\therefore \quad 2 p=4 \Rightarrow p=2\)
Since \((2,3)\) lies on \(y^2=p x^3+q\).
\(\therefore \quad 9=2 \times 8+q \Rightarrow q=-7\)
Differentiating both sides w.r.t. \(x\), we get
\(\begin{aligned}
& 2 y \cdot \frac{\mathrm{~d} y}{\mathrm{~d} x}=3 \mathrm{p} x^2 \\
& \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{3 \mathrm{p}}{2}\left(\frac{x^2}{y}\right) \\
\therefore \quad & \left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{(2,3)}=\frac{3 \mathrm{p}}{2} \times \frac{4}{3}=2 \mathrm{p}
\end{aligned}\)
Since the line touches the curve, their slopes are equal.
\(\therefore \quad 2 p=4 \Rightarrow p=2\)
Since \((2,3)\) lies on \(y^2=p x^3+q\).
\(\therefore \quad 9=2 \times 8+q \Rightarrow q=-7\)
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