MHT CET · Maths · Application of Derivatives
If the curves \(y^2=6 x\) and \(9 x^2+b y^2=16\) intersect each other at right angle, then value of " \(b\) ' is
- A \(\frac{9}{2}\)
- B \(6\)
- C \(\frac{7}{2}\)
- D \(4\)
Answer & Solution
Correct Answer
(A) \(\frac{9}{2}\)
Step-by-step Solution
Detailed explanation
\(y^2=6 x \)
\( \Rightarrow 2 y \frac{\mathrm{d} y}{\mathrm{~d} x}=6 \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3}{y} \)
\( \text {Also, } 9 x^2+\mathrm{b} y^2=16 \)
\( \Rightarrow 18 x+2 \mathrm{~b} y \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-9 x}{\mathrm{~b} y}\)
As given curves intersect each other at right angle, their tangents also intersect at right angles.
\(\frac{3}{y} \times \frac{-9 x}{\mathrm{~b} y}=-1 \)
\( \Rightarrow \mathrm{b} y^2=27 x \)
\( \therefore (\mathrm{i}) \Rightarrow \mathrm{b}(6 x)=27 x \)
\( \Rightarrow \mathrm{b}=\frac{9}{2}\)
\( \Rightarrow 2 y \frac{\mathrm{d} y}{\mathrm{~d} x}=6 \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3}{y} \)
\( \text {Also, } 9 x^2+\mathrm{b} y^2=16 \)
\( \Rightarrow 18 x+2 \mathrm{~b} y \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-9 x}{\mathrm{~b} y}\)
As given curves intersect each other at right angle, their tangents also intersect at right angles.
\(\frac{3}{y} \times \frac{-9 x}{\mathrm{~b} y}=-1 \)
\( \Rightarrow \mathrm{b} y^2=27 x \)
\( \therefore (\mathrm{i}) \Rightarrow \mathrm{b}(6 x)=27 x \)
\( \Rightarrow \mathrm{b}=\frac{9}{2}\)
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