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Question

# Find the points on the curve y = 3x2 − 9x + 8 at which the tangents are equally inclined with the axes.

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Solution

## Let (x1, y1) be the required point. It is given that the tangent at this point is equally inclined to the axes. It means that the angle made by the tangent with the x-axis is $±$45$°$. ∴ Slope of the tangent = tan ($±$45) = $±$ 1 ...(1) $\mathrm{Since},\mathrm{the}\mathrm{point}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{curve}.\phantom{\rule{0ex}{0ex}}\mathrm{Hence},{y}_{1}=3{{x}_{1}}^{2}-9{x}_{1}+8\phantom{\rule{0ex}{0ex}}\mathrm{Now},y=3{x}^{2}-9x+8\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=6x-9\phantom{\rule{0ex}{0ex}}\text{Slope of the tangent at}\left({x}_{1},{y}_{1}\right)\text{=}{\left(\frac{dy}{dx}\right)}_{\left({x}_{1},{y}_{1}\right)}\text{=6}{\mathit{\text{x}}}_{\mathit{1}}\text{-9 ...(2)}\phantom{\rule{0ex}{0ex}}\text{From eq. (1) and eq. (2), we get}\phantom{\rule{0ex}{0ex}}6{x}_{1}-9=±1\phantom{\rule{0ex}{0ex}}⇒6{x}_{1}-9=1\text{or}6{x}_{1}-9=-1\phantom{\rule{0ex}{0ex}}⇒6{x}_{1}=10\text{or}6{x}_{1}=8\phantom{\rule{0ex}{0ex}}⇒{x}_{1}=\frac{10}{6}=\frac{5}{3}\text{or}{x}_{1}=\frac{8}{6}=\frac{4}{3}\phantom{\rule{0ex}{0ex}}\mathrm{Also},\phantom{\rule{0ex}{0ex}}{y}_{1}=3{\left(\frac{5}{3}\right)}^{2}-9\left(\frac{5}{3}\right)+8\text{or}{y}_{1}=3{\left(\frac{4}{3}\right)}^{2}-9\left(\frac{4}{3}\right)+8\phantom{\rule{0ex}{0ex}}⇒{y}_{1}=\frac{25}{3}-\frac{45}{3}+8\text{or}{y}_{1}=\frac{16}{3}-\frac{36}{3}+8\phantom{\rule{0ex}{0ex}}⇒{y}_{1}=\frac{4}{3}\text{or}{y}_{1}=\frac{4}{3}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{required}\text{points are}\left(\frac{5}{3},\frac{4}{3}\right)\text{and}\left(\frac{4}{3},\frac{4}{3}\right)\text{.}$

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