Tangent to the Curve at a Point - Practice Questions & MCQ

Tangent and Normal

Slope and Equation of Tangent

Let $P\left(x_0, y_0\right)$ be a point on the continuous curve $y=f(x)$, then the slope of the tangent to the curve at point $P$ is

$
\begin{aligned}
& \left(\frac{d y}{d x}\right)_{\left(x_0, y_0\right)} \\
& \Rightarrow\left(\frac{d y}{d x}\right)_{\left(x_{0, y 0)}\right.}=\tan \theta=\text { slope of tangent at } P
\end{aligned}
$
Where $\theta$ is the angle which the tangent at $\mathrm{P}\left(\mathrm{x}_0, \mathrm{y}_0\right)$ makes with the positive direction of the x-axis as shown in the figure.

• If the tangent is parallel to $x$-axis then $\theta=0^{\circ}$.

  $
  \begin{aligned}
  & \Rightarrow \quad \tan \theta=0 \\
  & \therefore \quad\left(\frac{d y}{d x}\right)_{(x 0, y 0)}=0
  \end{aligned}
  $

  If the tangent is perpendicular to $x$-axis then $\Theta=90^{\circ}$

  $
  \begin{aligned}
  & \Rightarrow \quad \tan \theta \rightarrow \infty \quad \text { or } \quad \cot \theta=0 \\
  & \therefore \quad\left(\frac{d x}{d y}\right)_{\left(x_0, y_0\right)}=0
  \end{aligned}
  $
  Equation of Tangent
  Let the equation of curve be $y=f(x)$ and let point $P\left(x_0, y_0\right)$ lie on this curve.
  The slope of the tangent to the curve at a point $P$ is

  $
  \left(\frac{d y}{d x}\right)_{\left(x_0, y_0\right)}
  $
  Hence, the equation of the tangent at point $P$ is

  $
  \left(y-y_0\right)=\left(\frac{d y}{d x}\right)_{\left(x_0, y_0\right)} \cdot\left(x-x_0\right)
  $

Tangent from External Point

If a point $Q(a, b)$ does not lie on the curve $y=f(x)$, then the equation of possible tangent to the curve $y=f(x)$ (tangent passing through point $Q(a, b)$ ) can be found by first getting the point of contact $P\left(x_0\right.$, $\left.y_0\right)$ on the curve.

$P\left(x_0, y_0\right)$ lies on the curve $\mathrm{y}=\mathrm{f}(\mathrm{x})$, then

$
y_0=f\left(x_0\right)
$
Also, the slope of PQ is

$
\frac{y_0-b}{x_0-a}=\left(\frac{d y}{d x}\right)_{\left(x_0, y_0\right)}
$
By solving the above two equations we get the point of contact $P$.
Using P we can find the equation of tangent PQ.

Slope and Equation of Normal

The normal to a curve at a given point say $P\left(x_0, y_0\right)$ is a line perpendicular to the tangent at $P$ and which passes through $P$.
So, let the continuous curve be $y=f(x)$ and let the point $P\left(x_0, y_0\right)$ lies on this curve.
Therefore the slope of normal to the curve at point $\mathrm{P}\left(\mathrm{x}_0, \mathrm{y}_0\right)$ is given by :
Slope of normal at $\mathrm{P}=-\frac{1}{\text { slope of tangent at } \mathrm{P}}$

$
\begin{aligned}
& =-\frac{1}{\left(\frac{d y}{d x}\right)_{\left(x_0, y_0\right)}} \\
& =-\left(\frac{d x}{d y}\right)_{\left(x_0, y_0\right)} \\
 &
\end{aligned}
$

• - If normal is parallel to $x$-axis then

  $
  -\left(\frac{d x}{d y}\right)_{\left(x_0, y_0\right)}=0 \quad \text { or } \quad\left(\frac{d x}{d y}\right)_{\left(x_0, y_0\right)}=0
  $

  - If normal is perpendicular to x -axis then

  $
  -\left(\frac{d y}{d x}\right)_{\left(x 0, y_0\right)}=0 \quad \text { or } \quad\left(\frac{d y}{d x}\right)_{\left(x_0, y_0\right)}=0
  $
  The equation of the Normal at point $P\left(x_0, y_0\right)$ (which lies on the curve $\left.y=f(x)\right)$ is

  $
  \left(y-y_0\right)=-\left(\frac{d x}{d y}\right)_{\left(x_0, y_0\right)}\left(x-x_0\right)
  $
  Normal from External Point
  If point $Q(a, b)$ does not lie on the curve $y=f(x)$, then the equation of possible normal to the curve $y=f(x)$ passing through $Q$ is given by

  $
  \frac{y_0-b}{x_0-a}=-\left(\frac{d x}{d y}\right)_{\left(x_0, y_0\right)}
  $
  Where $\mathrm{P}\left(\mathrm{x}_0, \mathrm{y}_0\right)$ is the point where this normal cuts the curve
  Also $y_0=f\left(x_0\right)$
  We can solve these two equations to get point $P$ first, and then get the equation of normal.