Question

$y=x^2$ and $6y=7-x^2.$ make an angle of $\theta ={\tan }^{-1}\left(k\right)$ at points where they intersect. Find $k$?

Answer 1

Given curves
$y=x^2$ ....(1)
$6y=7-x^2$ .....(2)
Solving (1) and (2), we get
$x=\pm 1,y=1$
So, the point of intersection is (1,1) and (-1,1)
Now, differentiating (1) w.r.t. x, we get
$\frac{dy}{dx}=2x$
So, slope of tangent to curve (1) at (1,1) is 2.
And slope of tangent to curve (1) at (-1,1) is -2.
Now, differentiating (2) w.r.t. x, we get
$\frac{dy}{dx}=-\frac{1}{3}x$
So, slope of tangent to curve (1) at (1,1) is $-\frac{1}{3}$.
And slope of tangent to curve (1) at (-1,1) is $\frac{1}{3}$.
Angle between the tangents at $(1,1)$ is
$\tan \theta =\frac{m_1-m_2}{1+m_1{m}_2}$
$=\frac{2+\frac{1}{3}}{1-\frac{2}{3}}$
$\tan \theta =\frac{7}{1}$
$\Rightarrow \theta ={\tan }^{-1}7$
Angle between the tangents at $(-1,1)$ is
$\tan \theta =\frac{m_1-m_2}{1+m_1{m}_2}$
$=\frac{-2-\frac{1}{3}}{1-\frac{2}{3}}$
$\tan \theta =\frac{-7}{1}$
$\Rightarrow \theta ={\tan }^{-1}(-7)$
We know that angle between curves is equal to the angle between the tangents.
Hence, on comparing with given value
$k=\pm 7$