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 sub-tangent drawn to the curve at $P$ is equal to the length of the subnormal, then the area of the $△PAB$(in $squnits$) is

• A. $4$
• B. $32$
• C. $8$
• D. $16$

Answer 1

The correct option is D

$16$

Explanation for the correct option:

Step-1: Solve for the equations of tangent and normal

The equation of the curve is $y=x^2-x+4$

The first order derivative of the curve with respect to $x$ at a point gives the slope of the tangent at that point

Differentiating with respect to $x$ we get

$\Rightarrow$$\frac{dy}{dx}=2x-1$

$\Rightarrow$${\frac{dy}{dx}}_{\left(1,4\right)}=2\left(1\right)-1=1$

The equation of the tangent is given as

$y-y_1={\frac{dy}{dx}}_{\left(x_1,y_1\right)}\left(x-x_1\right)$

Given that tangent is at $P(1,4)$.

Substituting the values of the co-ordinates we get

$\Rightarrow$$y-4=1\left(x-1\right)$

$\Rightarrow$$x-y=-3$$...\left(i\right)$

The equation of the normal is given as

$y-y_1=\frac{-1}{{\frac{dy}{dx}}_{\left(x_1,y_1\right)}}\left(x-x_1\right)$

Given that normal is at $P(1,4)$. Substituting the values of the co-ordinates we get

$\Rightarrow$$y-4=-1\left(x-1\right)$

$\Rightarrow$$x+y=5$$...\left(ii\right)$

Step 2: Solve for the required area of the triangle

The tangent and the normal intersect the $x-$axis say at points $A$ and $B$ respectively.

On the $x-$axis , $y=0$

Substituting $y=0$ in equation of tangent we get $x=-3$

Hence, $A=\left(-3,0\right)$

Substituting $y=0$ in equation of normal we get $x=5$

Hence, $B=\left(5,0\right)$

Let $P(1,4)=\left(x_1,y_1\right)$, $A\left(-3,0\right)=\left(x_2,y_2\right)$ and $B\left(5,0\right)=\left(x_3,y_3\right)$

The area of the triangle formed by the vertices $\left(x_1,y_1\right)$, $\left(x_2,y_2\right)$ and $\left(x_3,y_3\right)$is given as

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

Substituting the values of the co-ordinates we get,

Area of the $△PAB$$=\frac{1}{2}\left|\begin{array}{ccc}1& 4& 1\\ -3& 0& 1\\ 5& 0& 1\end{array}\right|$

$=\frac{1}{2}\left[1\left(0\right)-4\left(-3-5\right)+1\left(0\right)\right]$

$=\frac{1}{2}\times 32$

$\Rightarrow$ Area of the $△PAB$$=16sq.units$

Hence, option(D) i.e. $16$ is the correct option.