Question

The curve y = (1 + x)(3 − x) intersects the positive side of the x-axis at A. The
tangent to the curve at O, the origin, intersects the curve again at P. Calculate
i. the finite area bounded by the curve and OA,
ii. the finite area bounded by the curve and OP

Answer 1

i

We are given the curve $y = (1 + x)(3 — x)$ , which intersects the positive side of the x-axis at A. The points of intersection of the curve with the x-axis are points where the y coordinate becomes 0, so we have 

$y= 0 \implies (1+ x)(3— x) = 0 \implies x= —1,3$

Hence the point A has the coordinates $(3, 0)$ . So, we have to find the area bounded by the curve y and OA, where 0 is the origin. So, we just have to find the area under the curve $y = (1+x)(3—x) = 3+2x—x^2$ between $x = 0$ and $x = 3$ . This is given by 

$Area=\int_0^3 ydx =\int_0^3 (3+2x-x^2)dx = [3x+x^2- \frac{x^3}{3}]_0^3=9+9-9=9 sq \space units$

ii

Now, we first have to find the tangent to the curve at 0, the origin. The slope of the tangent is given by the value of the derivative of y at x = 0, so we have 

$\frac{dy}{ dx} |_{x=0}=[\frac{d}{dx}(3+2x-x^2)]_{x=0}= [2 — 2x]_{x=0}= 2$

Also, the value of y at the origin is $y = (1 + 0)(3 — 0) = 3$ . So the equation of the tangent is given by

$\frac{y — 3}{x-0}= \frac{ dy}{dx}|_{x=0} \implies y= 2x + 3$

Now, P is the point of intersection of the above line with the curve. So, equation the two expressions we get

$(1 + x)(3 — x) = 2x + 3 \implies 3 + 2x -x^2 = 2x + 3 \implies x^2 = 0 \implies x = 0$

The above solution tells us that the tangent to the curve y intersects it at only one point, x = 0, which is the origin. So the point of intersection P does not exist. Hence the area bounded by the curve and OP does not exist.