MHT CET · Maths · Application of Derivatives
If the curves \(y^2=6 x, 9 x^2+b y^2=16\) intersect each other at right angles, the value of \(b\) is
- A 4
- B \(\frac{7}{2}\)
- C 6
- D \(\frac{9}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{9}{2}\)
Step-by-step Solution
Detailed explanation
\(\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{C_1}=\frac{3}{y} \text { and }\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{C_2}=\frac{-9 x}{b y}\)
for orthogonal intersection \(\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{C_1} \times\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{C_2}=-1\)
\(\begin{aligned}
& \Rightarrow \frac{3}{y} \times \frac{-9 x}{b y}=-1 \\
& \Rightarrow 27 x=b y^2 \\
& \Rightarrow 27 x=b \times 6 x\left[\because y^2=6 x\right] \\
& \Rightarrow b=\frac{9}{2}
\end{aligned}\)
for orthogonal intersection \(\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{C_1} \times\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{C_2}=-1\)
\(\begin{aligned}
& \Rightarrow \frac{3}{y} \times \frac{-9 x}{b y}=-1 \\
& \Rightarrow 27 x=b y^2 \\
& \Rightarrow 27 x=b \times 6 x\left[\because y^2=6 x\right] \\
& \Rightarrow b=\frac{9}{2}
\end{aligned}\)
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