Question

The tangent lines to the curve $y^2=4ax$ at points where $x=a$, are

• A. parallel
• B. perpendicular
• C. inclined at $60^{\circ }$
• D. inclined at $30^{\circ }$

Answer 1

Answer: (B)

The given equation of the curve is $y^2=4ax...(1)$
Differentiating both sides of $(1)$ with respect to $x$, we get
$2y\frac{dy}{dx}=4a;$
$\Rightarrow \frac{dy}{dx}=\frac{4a}{2y}=\frac{2a}{y}...(2)$
If $\psi$ be the angle which the tangent to the curve at $(x,y)$
makes with the positive direction of $x$-axis then tan
$\psi =\frac{dy}{dx}$ or $\tan \psi =$
$\frac{2a}{y}...(3),$[using (2)]
At $x=a,$ then from (1),
$y^2=4a\cdot a=4a^2$
$\Rightarrow y=\pm 2a$
Hence, we get two points $(a,2a)$ and $(a,-2a)$ on the
curve.
At $(a,2a)x=a,y=2a$ and let $\psi ={\psi }_1$
$\therefore$ from $(3),\tan {\psi }_1=\frac{2a}{2a}=1=\tan 45^{\circ };$ $\Rightarrow {\psi }_1=45^{\circ }$
At $(a,-2a),x=a,y=-2a$ and let $\psi ={\psi }_2.$
$\therefore$ from $(3),\tan {\psi }_2=\frac{2a}{-2a}=-1=\tan 135^{\circ };$
or ${\psi }_2$
$=135^{\circ }.$
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 the tangent lines to $(1)$ at $(a,2a)$ and $(a,-2a)$ are perpendicular to each other.