1 \(\pm \tan ^{-1}\left(\frac{4}{29}\right)\)
2 \(\pm \tan ^{-1}\left(\frac{5}{29}\right)\)
3 \(\pm \tan ^{-1}\left(\frac{10}{49}\right)\)
4 \(\pm \tan ^{-1}\left(\frac{8}{29}\right)\)
Explanation:
(C) : Equation of the given curve in parametric form,
\(\mathrm{x}=\mathrm{t}^{2}+1 \text { and } \mathrm{y}=\mathrm{t}^{2}-\mathrm{t}-6\)
\(\mathrm{Y}\)-coordinate of the point, where the given curve meets \(\mathrm{X}\)-axis is 0 .
When, \(\mathrm{y}=0\), then \(\mathrm{t}^{2}-\mathrm{t}-6=0\)
\(t^{2}-3 t+2 t-6=0\)
\(t(t-3)+2(t-3)=0\)
\((t-3)(t+2)=0\)
\(t=3 \text { or }-2\)
When, \(\mathrm{t}=3\), then \(\mathrm{x}=10\)
when, \(t=-2\), then \(x=5\)
Hence, the points where the curve meets the \(\mathrm{X}\)-axis are \((10,0)\) and \((5,0)\).
Now, \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}=\frac{2 \mathrm{t}-1}{2 \mathrm{t}}\)
Slope of the tangent at point \((10,0)\)
\(\mathrm{m}_{1}=\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{x}=10}=\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{t}=3}=\frac{5}{6}\)
Slope of the tangent at point \((5,0)\),
\(\mathrm{m}_{2}=\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{x}=5}=\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{t}=-2}=\frac{-5}{-4}=\frac{5}{4}\)
If \(\theta\) be the angle between two tangents, then
\(\tan \theta=\left|\frac{\mathrm{m}_{2}-\mathrm{m}_{1}}{1+\mathrm{m}_{1} \times \mathrm{m}_{2}}\right|=\left|\frac{\frac{5}{4}-\frac{5}{6}}{1+\frac{5}{6} \cdot \frac{5}{4}}\right|\)
\(=\left|\frac{\frac{15-10}{12}}{\frac{24+25}{24}}\right|=\left|\frac{\frac{5}{12}}{\frac{49}{24}}\right|=\left|\frac{10}{49}\right|\)
\(\tan \theta= \pm \frac{10}{49}\)
Therefore, \(\theta=\tan ^{-1}\left( \pm \frac{10}{49}\right)= \pm \tan ^{-1}\left(\frac{10}{49}\right)\)
