MHT CET · Maths · Application of Derivatives
The angle between the tangents to the curves \(y=2 x^2\) and \(x=2 y^2\) at \((1,1)\) is
- A \(\tan ^{-1}\left(\frac{15}{8}\right)\)
- B \(\tan ^{-1}\left(\frac{7}{8}\right)\)
- C \(\tan ^{-1}\left(\frac{3}{4}\right)\)
- D \(\tan ^{-1}\left(\frac{1}{4}\right)\)
Answer & Solution
Correct Answer
(A) \(\tan ^{-1}\left(\frac{15}{8}\right)\)
Step-by-step Solution
Detailed explanation
\(
y=2 x^2
\)
\(\therefore\) Slope of the tangent to this curve is \(\frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{m}_1=4 x\)
\(\therefore \) at \((1,1), \mathrm{m}_1=4\)
\(
x=2 y^2
\)
\(\therefore \) Slope of the tangent to this curve is \(\frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{m}_2=\frac{1}{4 y}\)
\(\therefore \) at \((1,1), \mathrm{m}_2=\frac{1}{4}\)
Let \(\theta\) be the angle between two tangents.
\(\therefore \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{4-\frac{1}{4}}{1+4 \times \frac{1}{4}}\right|=\frac{15}{8}\)
\(\therefore \theta=\tan ^{-1}\left(\frac{15}{8}\right)\)
y=2 x^2
\)
\(\therefore\) Slope of the tangent to this curve is \(\frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{m}_1=4 x\)
\(\therefore \) at \((1,1), \mathrm{m}_1=4\)
\(
x=2 y^2
\)
\(\therefore \) Slope of the tangent to this curve is \(\frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{m}_2=\frac{1}{4 y}\)
\(\therefore \) at \((1,1), \mathrm{m}_2=\frac{1}{4}\)
Let \(\theta\) be the angle between two tangents.
\(\therefore \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{4-\frac{1}{4}}{1+4 \times \frac{1}{4}}\right|=\frac{15}{8}\)
\(\therefore \theta=\tan ^{-1}\left(\frac{15}{8}\right)\)
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