MHT-CET Full Test-8 Mathematics Que-12 Solution

Q.12.The angle between the tangents to the curves y=2x2 and x=2y2 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 :- A. tan^{-1}\left ( \frac{15}{8} \right )

Explanation :-

y=2x2

Slope of the tangent to this curve is

\frac{dy}{dx}=m_{1}=4x

at (1,1), m1=4

x=2y2

Slope of the tangent to this curve is \frac{dy}{dx}=m_{2}=\frac{1}{4y}

at (1,1), m2= \frac{1}{4}

Let \theta be the angle between two tangents

tan\; \theta =\left | \frac{m_{1}-m_{2}}{1+m_{1}m_{2}}\right | tan\; \theta =\left | \frac{4-\frac{1}{4}}{1+4\times \frac{1}{4}}\right |=\frac{15}{8} \theta =tan^{-1}\left ( \frac{15}{8} \right )
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