Question

Which one of the following curves cuts the parabola $y^2=4ax$ at right angles.

• A. $x^2+y^2=a^2$
• B. $y=e^{\frac{-x}{2a}}$
• C. $y=ax$
• D. $x^2=4ay$

Answer 1

The correct option is B

$y=e^{\frac{-x}{2a}}$

Explanation for the correct option:

Option(B): Given that equation of parabola is $y^2=4ax$

Slope of a curve is given by $\frac{dy}{dx}$.

Therefore differentiating both sides of the equation with respect to $x$ we get,

$\begin{array}{crcl}& \frac{d}{dx}\left(y^2\right)& =& \frac{d}{dx}\left(4ax\right)\\ \Rightarrow & 2y\frac{dy}{dx}& =& 4a\\ \Rightarrow & \frac{dy}{dx}& =& \frac{4a}{2y}\end{array}$

Therefore slope of the given parabola is $m_1=\frac{4a}{2y}=\frac{2a}{y}$

Let the slope of curve that cuts the given parabola be $m_2$

For the two curves to intersect at right angle, the condition is $m_1·m_2=-1$

Now consider the equation of the curve $y=e^{\frac{-x}{2a}}$

Now differentiating on both sides with respect to $x$, we get,

$\begin{array}{crcll}& \frac{d}{dx}\left(y\right)& =& \frac{d}{dx}\left(e^{\frac{-x}{2a}}\right)& \\ \Rightarrow & \frac{dy}{dx}& =& \frac{-1}{2a}\left(e^{\frac{-x}{2a}}\right)& \left[\because \frac{d}{dx}\left[e^{ax}\right]=a{e}^{ax}\right]\\ \Rightarrow & \frac{dy}{dx}& =& \frac{-1}{2a}\left(y\right)& \left[\because y=e^{\frac{-x}{2a}}\right]\\ \Rightarrow & m_2& =& \frac{-y}{2a}& \end{array}$

Now $m_1·m_2=\frac{2a}{y}\times \frac{-y}{2a}=-1$i.e. both curves intersect at right angle.

Thus, option(B) is correct.

Explanation for incorrect options:

Option(A): Now consider the equation of the curve $x^2+y^2=a^2$

Now differentiating on both sides with respect to $x$, we get,

$\begin{array}{crcl}& \frac{d}{dx}\left(x^2+y^2\right)& =& \frac{d}{dx}\left(a^2\right)\\ \Rightarrow & 2x+2y\frac{dy}{dx}& =& 0\\ \Rightarrow & \frac{dy}{dx}& =& \frac{-2x}{2y}\\ \Rightarrow & m_2& =& \frac{-x}{y}\end{array}$

Now $m_1·m_2=\frac{2a}{y}\times \frac{-x}{y}\ne -1$i.e. both curves don't intersect at right angle.

Thus, option(A) is incorrect.

Option(C): Now consider the equation of the curve $y=ax$

Now differentiating on both sides with respect to $x$, we get,

$\begin{array}{crcl}& \frac{d}{dx}\left(y\right)& =& \frac{d}{dx}\left(ax\right)\\ \Rightarrow & \frac{dy}{dx}& =& a\\ \Rightarrow & m_2& =& a\end{array}$

Now $m_1·m_2=\frac{2a}{y}\times a\ne -1$ i.e. both curves don't intersect at right angle.

Thus, option(C) is incorrect.

Option(D): Now consider the equation of the curve $x^2=4ay$

Now differentiating on both sides with respect to $x$, we get,

$\begin{array}{crcl}& \frac{d}{dx}\left(x^2\right)& =& \frac{d}{dx}\left(4ay\right)\\ \Rightarrow & 2x& =& 4a\frac{dy}{dx}\\ \Rightarrow & \frac{dy}{dx}& =& \frac{x}{2a}\\ \Rightarrow & m_2& =& \frac{x}{2a}\end{array}$

Now $m_1·m_2=\frac{2a}{y}\times \frac{x}{2a}\ne -1$ i.e. both curves don't intersect at right angle.

Hence the correct option is option(B) i.e. $y=e^{\frac{-x}{2a}}$