The tangent at the point \((1,2)\) to the curve \(y^2=4 x\) makes an angle \(\theta\) with the positive…
The tangent at the point \((1,2)\) to the curve \(y^2=4 x\) makes an angle \(\theta\) with the positive direction of \(X\)-axis. Then \(\theta=\)
- \(60^{\circ}\)
- \(30^{\circ}\)
- \(90^{\circ}\)
- \(45^{\circ}\)
Solution
\(y^2=4 x ; P=(1,2)\)
differentiate w.r. to ' \(x\) '
\(\begin{aligned}
2 y \cdot y^{\prime} & =4 \cdot 1 \\
y^{\prime} & =\frac{4}{2 y} \\
y^{\prime} & =\frac{2}{y}
\end{aligned}\)
\(\begin{aligned}
\text {Slope of tangent } & =\left.y^{\prime}\right|_{P(1,2)}=\left.\frac{2}{y}\right|_{P(1,2)} \\
\tan \theta & =1 \Rightarrow \theta=45^{\circ}
\end{aligned}\)
Hence, option (d) is correct.
Asked in: AP EAMCET 2020 (18 Sep Shift 2)
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