Question

The tangent and the normal drawn to the curve $y=x^2-x+4$ at $P(1,4)$ cut the $x$– axis at $A$ and $B$ respectively. If the length of the subtangent drawn to the curve at $P$ is equal to the length of the subnormal then the area of the triangle $PAB$ in sq. units is

• A. $4$
• B. $32$
• C. $8$
• D. $16$

Answer 1

The correct option is D

$16$

Explanation for the correct option:

Step 1: Calculate the slope tangent

Given curve is, $y=x^2-x+4$

The tangent to a curve can be found by calculating the derivative of the curve. Thus, the tangent of the given curve is,

$\begin{array}{rcl}\frac{\mathrm{dy}}{\mathrm{d}x}& =& \frac{\mathrm{d}}{\mathrm{d}x}\left(x^2-x+4\right)\\ & =& 2x-1\end{array}$

The slope of the tangent at a given point can be calculated by substituting the value of $x$-coordinate of the point into the equation of the tangent. So, the slope of the tangent at $P(1,4)$ is,

${\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}_{\left(1,4\right)}=2\times 1-1=1$

Step 2: Determine the equation of the tangent

The equation of the tangent to the line can be determined using the two-point form of a line which is given by,

$\left(y-y_1\right)=m\left(x-x_1\right)$

Here, $\left(x_1,y_1\right)=\left(1,4\right)$ and the slope $m=1$

Thus, the equation of the tangent line is,

$\begin{array}{cc}& y-4=1\left(x-1\right)\\ \Rightarrow & y=x+3\end{array}$

Step 3: Calculate the slope of the normal

The normal to a curve at a point is the line that is perpendicular to the tangent to the curve at that point.

We know that two lines are perpendicular if the sum of their product is $-1$.

Thus, the normal to a point of a curve is the negative reciprocal of the tangent, i.e.,

$-\frac{1}{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}=-\frac{1}{2x-1}$

The slope of the normal is calculated the same way as that of a tangent.

Therefore, the slope of the normal to the curve at $P(1,4)$ is,

$-\frac{1}{2\times 1-1}=-1$

Step 4: Determine the equation of the normal to the curve

The equation of the normal line can be determined using the two-point form of a line which is given by,

$\left(y-y_1\right)=m\left(x-x_1\right)$

Here, $\left(x_1,y_1\right)=\left(1,4\right)$ and the slope $m=-1$

Thus, the equation of the tangent line is,

$\begin{array}{cc}& y-4=-1\left(x-1\right)\\ \Rightarrow & y=5-x\end{array}$

Step 5: Determine the coordinates of point $A$

The equation of the tangent to the curve is $y=x+3$ (as derived earlier)

Given, point $A$ is the $x$-intercept of the tangent.

This implies that the $y$-coordinate of the point $A$ is $0$.

Substituting this in the equation of the tangent, we get

$\begin{array}{cc}& 0=x+3\\ \Rightarrow & x=-3\end{array}$

Therefore, the coordinates of the point $A$ is $\left(-3,0\right)$.

Step 6: Determine the coordinates of point $B$

The equation of the tangent to the curve is $y=5-x$ (as derived earlier)

Given, point $B$ is the $x$-intercept of the tangent.

This implies that the $y$-coordinate of the point $B$ is $0$.

Substituting this in the equation of the tangent, we get

$\begin{array}{cc}& 0=5-x\\ \Rightarrow & x=5\end{array}$

Therefore, the coordinates of the point $A$ is $\left(5,0\right)$.

Step 7: Use the determinant method to find the area of the triangle

The coordinates of $△PAB$ are $\left(1,4\right)$, $\left(-3,0\right)$ and $\left(5,0\right)$ of points $P$, $A$ and $B$ respectively.

The area of a triangle when the coordinates of its three vertices are given is given by the formula,

$A=\pm \frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

where the first column is the $x$ coordinates of the vertices and the second column is their respective $y$ coordinates.

Thus, the area of $△PAB$ is,

$\begin{array}{rcl}A& =& \pm \frac{1}{2}\left|\begin{array}{ccc}1& 4& 1\\ -3& 0& 1\\ 5& 0& 1\end{array}\right|\\ & =& \pm \frac{1}{2}\left[1\left(0-0\right)-4\left(-3\times 1-1\times 5\right)+1\left(0-0\right)\right]\\ & =& \pm \frac{1}{2}\left[-4\left(-3-5\right)\right]\\ & =& \pm \frac{1}{2}\left(32\right)\\ & =& 16\end{array}$

Therefore, the area of the $△PAB$ is $16$ sq.units.

Hence, option D is the correct option.