Explanation:
(B) : Given,
\(y^{2}=4 a x \tag{i}\)
On differentiating both sides \(\mathrm{w} \cdot \mathrm{r} \cdot \mathrm{t} \cdot \mathrm{x}\), we get-
\(\begin{aligned} & 2 y \frac{d y}{d x}=4 a \Rightarrow \frac{d y}{d x}=\frac{4 a}{2 y} \\ & \frac{d y}{d x}=\frac{2 a}{y}\end{aligned}\)
If \(\psi\) be the angle of the tangent to the curve at \((x, y)\) makes with the positive direction of \(\mathrm{x}\) - axis
Then, \(\quad \tan \psi=\frac{\mathrm{dy}}{\mathrm{dx}}\)
On putting the value \(\frac{d y}{d x}\) in above equation
\(\tan \psi=\frac{2 \mathrm{a}}{\mathrm{y}} \tag{iii}\)
At, \(\mathrm{x}=\mathrm{a}\), then from equation (i)
\(y^{2}=4 \mathrm{a} \cdot \mathrm{a}\)
\(\mathrm{y}^{2}=4 \mathrm{a}^{2}\)
\(\mathrm{y}= \pm 2\)
Hence, we get two points \((a, 2 a)\) and \((a,-2 a)\) on the curve.
At, (a, 2a)
\(\mathrm{x}=\mathrm{a}, \mathrm{y}=2 \mathrm{a}\) and Let \(\psi=\psi_{1}\)
Therefore, From equation (iii),
\(\tan \psi_{1}=\frac{2 \mathrm{a}}{2 \mathrm{a}}=1\)
\(\tan \psi_{1}=\tan 45\)
\(\psi_{1}=45^{\circ} .\)
\(\text { At }(\mathrm{a},-2 \mathrm{a}),\)
\(\mathrm{x}=\mathrm{a}, \mathrm{y}=-2 \mathrm{a}\)
Let, \(\psi_{1}=\psi_{2}\).
\(\therefore\) Form equation (iii) we have,
\(\tan \psi_{2} =\frac{2 \mathrm{a}}{-2 \mathrm{a}}\)
\(\tan \psi_{2} =-1\)
\(\tan \psi_{2} =\tan 135^{\circ}\)
\(\psi_{2} =135^{\circ}\)
Hence, the required angle between tangents to
(1) at (a, 2a) and (a,-2a)= \(\psi_{2}-\psi_{1}\)
\(135^{\circ}-45^{\circ}=90^{\circ}\)
This shows that tangent lines to (1) at (a, 2a) and (a, 2a) are perpendicular to each other.
