# Calculus Exercise (4) : Tangent Line and Normal Line to a Curve

1. Find the gradient of the curve $y=3 x^{2}-4 x+3$ at the point where $x=2$.

2. \begin{aligned} y&=3 x^{2}-4 x+3 \\\\ \frac{d y}{d x}&=6 x-4 \\\\ \left.\frac{d y}{d x}\right|_{x=2}&=6(2)-4\\\\ &=8\\\\ \end{aligned}
Hence, the gradient of the curve at $x=2$ is 8 .

3. Given that the gradient of the curve $y=x^{2}+a x+b$ at the point $(2,-1)$ is $1$. Find the values of $a$ and $b$.

4. \begin{aligned} y&=x^{2}+a x+b \\\\ \text { At }(2,-1),-1&=(2)^{2}+a(2)+b \\\\ 2 a+b&=-5 \\\\ b&=-5-2 a \\\\ \frac{d y}{d x}&=2 x+a\\\\ \left.\frac{d y}{d x}\right|_{(2,-1)}&=4+a\\\\ \left.\frac{d y}{d x}\right|_{(2,-1)}&=1\quad \text { (given) } \\\\ \therefore\ 4+a&=1\\\\ a&=-3 \\\\ \therefore\ b&=-5-2(-3)\\\\&=1 \end{aligned}

5. Find an equation of the tangent line and normal line to the graph of the function at the given point.

$\begin{array}{lll} {} & \textbf{Function} & \textbf{Point}\\\\ \text{(a)} & y=-2 x^{4}+5 x^{2}-3 & (1,0)\\\\ \text{(b)} & y=x^{3}-3 x & (2,2)\\\\ \text{(c)} & y=(x-2)\left(x^{2}+3 x\right)& (1,-4)\\\\ \text{(d)} & y=\dfrac{2}{x^{\frac{3}{4}}} & (1,2)\\ \end{array}$

6. \begin{aligned} \text{(a)} \quad \text{ Curve}: y&=-2 x^{4}+5 x^{2}-3\\\\ \text{Let} \left(x_{1}, y_{1}\right)&=(1,0)\\\\ \frac{d y}{d x}&=-8 x^{3}+10 x \\\\ \therefore\ m&=\left.\frac{d y}{d x}\right|_{(1,0)}\\\\ &=-8+10\\\\ &=2\\\\ \end{aligned}
The equation of tangent at $\left(x_{1}, y_{1}\right)$ is $y-y_{1}=m\left(x-x_{1}\right)$
\begin{aligned} &\\ \therefore\ y-0&=2(x-1) \\\\ \therefore\ 2 x-y&=2\\\\ \end{aligned}
The equation of normal at $\left(x_{1}, y_{1}\right)$ is
\begin{aligned} &\\ y-y_{1}&=-\frac{1}{m}\left(x-x_{1}\right) \\\\ \therefore\ y-0&=-\frac{1}{2}(x-1) \\\\ \therefore\ x+2 y&=1 \end{aligned}

\begin{aligned} \text{(b)} \quad \text{Curve}: y&=x^{3}-3 x\\\\ \text{Let} \left(x_{1}, y_{1}\right)&=(2,2) \frac{d y}{d x}&=3 x^{2}-3 \\\\ \therefore\ m&=\left.\frac{d y}{d x}\right|_{(2,2)}&=12-3\\\\ &=9\\\\ \end{aligned}
The equation of tangent at $\left(x_{1}, y_{1}\right)$ is
\begin{aligned} &\\ y-y_{1}&=m\left(x-x_{1}\right) \\\\ \therefore\ y-2&=9(x-2) \\\\ \therefore\ 9 x-y&=16\\\\ \end{aligned}
The equation of normal at $\left(x_{1}, y_{1}\right)$ is
\begin{aligned} &\\ y-y_{1}&=-\frac{1}{m}\left(x-x_{1}\right) \\\\ \therefore\ y-2&=-\frac{1}{9}(x-2) \\\\ \therefore\ x+9 y&=20 \end{aligned}

\begin{aligned} \text {(c)}\quad \text { Curve: } y&=(x-2)\left(x^{2}+3 x\right)&=x^{3}+x^{2}-6 x \\\\ \text { Let }\left(x_{1}, y_{1}\right)&=(1,-4) \\\\ \frac{d y}{d x}&=3 x^{2}+2 x-6 \\\\ \therefore\ m&=\left.\frac{d y}{d x}\right|_{(1,-4)}\\\\ &=3+2-6\\\\ &=-1 \\\\ \end{aligned}
The equation of tangent at \left(x_{1}, y_{1}\right) is
\begin{aligned} &\\ y-y_{1}&=m\left(x-x_{1}\right) \\\\ \therefore\ y+4&=(-1)(x-1) \\\\ \therefore\ x+y&=-3 \\\\ \end{aligned}
The equation of normal at \left(x_{1}, y_{1}\right) is
\begin{aligned} &\\ y-y_{1}&=-\frac{1}{m}\left(x-x_{1}\right) \\\\ \therefore\ y+4&=1(x-1) \\\\ \therefore\ x-y&=5 \end{aligned}

\begin{aligned} \text {(d)}\quad \text { Curve: } y&=\frac{2}{x^{\frac{3}{4}}} \\\\ \text { Let } \left(x_{1}, y_{1}\right)&=(1,2) \\\\ \frac{d y}{d x}&=-\frac{3}{2 x^{\frac{7}{4}}} \\\\ \therefore\ m&=\left.\frac{d y}{d x}\right|_{(1,2)}\\\\ &=-\frac{3}{2}\\\\ \end{aligned}
The equation of tangent at $\left(x_{1}, y_{1}\right)$ is
\begin{aligned} &\\ y-y_{1}&=m\left(x-x_{1}\right) \\\\ \therefore\ y-2&=\left(-\frac{3}{2}\right)(x-1) \\\\ \therefore\ 3 x+2 y&=7 \end{aligned}
The equation of normal at $\left(x_{1}, y_{1}\right)$ is
\begin{aligned} &\\ y-y_{1}&=-\frac{1}{m}\left(x-x_{1}\right) \\\\ \therefore\ y-2&=\frac{2}{3}(x-1) \\\\ \therefore\ 2 x-3 y&=-4 \end{aligned}

7. Find the equations of the tangent and normal lines to the curve $y=3 x^{2}-3 x+2$ at the point where $x=3$.

8. \begin{aligned} \text { Curve: } y=3 x^{2}-3 x+2 \\\\ \text { When } x=3, y=3(3)^{2}-3(3)+2=20 \\\\ \text{Let}\left(x_{1}, y_{1}\right)=(3,20) \\\\ \frac{d y}{d x}=6 x-3=3(2 x-1)\\\\ \therefore\ m=\left.\frac{d y}{d x}\right|_{(3,20)}=3(6-1)=15\\\\ \end{aligned}
The equation of tangent at $\left(x_{1}, y_{1}\right)$ is
\begin{aligned} &\\ y-y_{1}=m\left(x-x_{1}\right) \\\\ \therefore\ y-20=15(x-3) \\\\ \therefore\ 15 x-y=25\\\\ \end{aligned}
The equation of normal at $\left(x_{1}, y_{1}\right)$ is
\begin{aligned} &\\ y-y_{1}=-\frac{1}{m}\left(x-x_{1}\right) \\\\ \therefore\ y-20=-\frac{1}{15}(x-3) \\\\ \therefore\ x+15 y=303 \end{aligned}

9. Find the equation of the tangent to the curve $y=x^{2}+5 x-2$ at the point on the curve where this curve cuts the line $x=4$.

10. \begin{aligned} \text { Curve: } y &=x^{2}+5 x-2 \\\\ \text { When } x &=4, \\\\ y &=(4)^{2}+5(4)-2\\\\ &=34 \\\\ \text{Let} \left(x_{1}, y_{1}\right) &=(4,34) \\\\ \frac{d y}{d x} &=2 x+5\\\\ \therefore m &=\left.\frac{d y}{d x}\right|_{(4,34)}\\\\ &=2(4)+5\\\\ &=13\\\\ \end{aligned}
The equation of tangent at $\left(x_{1}, y_{1}\right)$ is
\begin{aligned} &\\ y-y_{1} &=m\left(x-x_{1}\right) \\\\ \therefore\ y-34 &=13(x-4) \\\\ \therefore\ 13 x-y &=18 \end{aligned}

11. Find the equations of the tangent and normal lines to the curve $y=x^{2}-5 x+6$ at the points where this curve cuts the $x$-axis.

12. Curve: $y=x^{2}-5 x+6\\\\$
When this curve cuts the $x$-axis, $y=0$
\begin{aligned} &\\ \therefore x^{2}-5 x+6 &=0 \\\\ (x-2)(x-3) &=0 \\\\ x &=2 \text { or } x &=3\\\\ \end{aligned}
Therefore, the curve cuts the $x$-axis at $(2,0)$ and $(3,0)\\\\$.
Let $\left(x_{1}, y_{1}\right) =(2,0)$ and $\left(x_{2}, y_{2}\right)=(3,0)$.
\begin{aligned} &\\ \therefore\ \frac{d y}{d x} &=2 x-5\\\\ \end{aligned}
Let the gradient of the curve at $(2,0)$ and $(3,0)$ be $m_{1}$ and $m_{2}$ respectively.
\begin{aligned} &\\ \therefore m_{1} &=\left.\frac{d y}{d x}\right|_{(2,0)} &=2(2)-5 &=-1 \\\\ \therefore m_{2} &=\left.\frac{d y}{d x}\right|_{(3,0)} &=2(3)-5 &=1\\\\ \end{aligned}
$\therefore$ The equation of tangent at $\left(x_{1}, y_{1}\right)$ is
\begin{aligned} &\\ y-y_{1} &=m_{1}\left(x-x_{1}\right) \\\\ y-0 &=-1(x-2) \\\\ y &=2-x\\\\ \end{aligned}
The equation of normal line at $\left(x_{1}, y_{1}\right)$ is
\begin{aligned} &\\ y-y_{1} &=-\frac{1}{m_{1}}\left(x-x_{1}\right) \\\\ y-0 &=1(x-2) \\\\ y &=x-2\\\\ \end{aligned}
The equation of tangent at $\left(x_{2}, y_{2}\right)$ is
\begin{aligned} &\\ y-y_{2} &=m_{2}\left(x-x_{2}\right) \\\\ y-0 &=1(x-3) \Rightarrow y &=x-3\\\\ \end{aligned}
The equation of normal line at $\left(x_{2}, y_{2}\right)$ is
\begin{aligned} & \\ y-y_{2} &=-\frac{1}{m_{2}}\left(x-x_{2}\right) \\\\ y-0 &=-1(x-3) \\\\ y &=3-x \end{aligned}

13. $P$ is the point $(3,4)$ on the curve $y=3 x^{2}-12 x+13$. Find the coordinates of the point of intersection of the normal to the curve at $P$ with the line $x+3=0$.

14. Curve: $y=3 x^{2}-12 x+13\\\\$,
Line: $x+3=0 \Rightarrow x=-3\\\\$
$P(3,4)$ is on the curve.
Let $\left(x_{1}, y_{1}\right)=(3,4)\\\\$.
$\therefore\ \dfrac{d y}{d x}=6 x-12=6(x-2)\\\\$
$\therefore\ m=\left.\dfrac{d y}{d x}\right|_{(3,4)}=6(3-2)=6\\\\$
The equation of normal line at $\left(x_{1}, y_{1}\right)$ is
\begin{aligned} &\\ y-y_{1}&=-\frac{1}{m}\left(x-x_{1}\right) \\\\ y-4&=-\frac{1}{6}(x-3) \\\\ y&=-\frac{1}{6}(x-3)+4\\\\ \end{aligned}
When this normal line intersects the line $x+3=0$, $x=-3$.
\begin{aligned} &\\ \therefore\ y&=-\frac{1}{6}(-3-3)+4\\\\ &=5\\\\ \end{aligned}
So, the point of intersection of the normal to the curve at $P(3,4)$ with the line $x+3=0$ is $(-3,5)$.

15. If the line $2 x+y=3$ is tangent to the curve $y=k x^{2}$, find the value of $k$.

16. \begin{aligned} \text{Curve }: y&=k x^{2},\\\\ \therefore\ \text{ Gradient of tangent } &=\frac{d y}{d x}\\\\ &=2 k x\\\\ \text{Tangent }: 2 x+y&=3 \\\\ y&=-2 x+3\\\\ \therefore\ \text{ Gradient of tangent } &=-2\\\\ \therefore\ 2 k x&=-2 \\\\ x&=-\frac{1}{k}\\\\ \end{aligned}
Substituting $x=-\dfrac{1}{k}$ in curve and line equations,
\begin{aligned} &\\ y&=k\left(-\frac{1}{k}\right)^{2} \\\\ y&=\frac{1}{k} \\\\ \therefore\ \frac{1}{k}&=-2\left(-\frac{1}{k}\right)+3\\\\ k&=-\frac{1}{3} \end{aligned}

17. Find an equation of the tangent line to the curve $y=x^{4}+1$ that is parallel to the line $32 x-y=15$.

18. \begin{aligned} \text { Curve: } y &=x^{4}+1 \\\\ \therefore \text { Gradient of tangent } &=\frac{d y}{d x} \\\\ &=4 x^{3} \\\\ \text { Line: } 32 x-y &=15 \\\\ y &=32 x-15 \\\\ \therefore\ \text { Gradient of line } &=32 \\\\ \end{aligned}
Since the tangent line to the curve is parallel to the line $32 x-y$,
\begin{aligned} &\\ 4 x^{3} &=32 \\\\ x &=2 \\\\ \text { When } x &=2, \\\\ y &=(2)^{4}+1 \\\\ &=17 \\\\ \text { Let }\left(x_{1}, y_{1}\right) &=(2,17). \\\\ \therefore\ \text { The equation of tangent at } & \left(x_{1}, y_{1}\right) \text { is } \\\\ y-y_{1} &=m\left(x-x_{1}\right) \\\\ \therefore\ y-17 &=32(x-2) \\\\ \therefore\ 32 x-y &=47 \end{aligned}

19. Find an equation of the normal line to the curve $y=\sqrt{x}$ that is parallel to the line $2 x+y=1$.

20. \begin{aligned} \text { Curve: } y &=\sqrt{x} \\\\ \therefore \text { Gradient of tangent } &=\frac{d y}{d x} \\\\ &=\frac{1}{2 \sqrt{x}} \\\\ \therefore \text { Gradient of normal } &=-2 \sqrt{x} \\\\ \text { Line }: 2 x+y &=1 \\\\ \qquad y &=-2 x+1 \\\\ \therefore \text { Gradient of line } &=-2 \\\\ \end{aligned}
Since the normal line to the curve is parallel to the line $2 x+y =1$.
\begin{aligned} &\\ -2 \sqrt{x}=-2 & \\\\ \qquad x=1 & & \\\\ \text { When } x=1, y=\sqrt{1}=1\\\\ \text { Let } \left(x_{1}, y_{1}\right)=(1,1)\\\\ \end{aligned}
$\therefore$ The equation of tangent at $\left(x_{1}, y_{1}\right)$ is
\begin{aligned} &\\ y-1&=-2(x-1) \\\\ 2 x+y&=3 \end{aligned}

21. Find an equation of the tangent line to the graph of $y=f(x)$ at $x=5$ if $f(5)=-3$ and $f^{\prime}(5)=4$.

22. \begin{aligned} &\text { Curve }: y=f(x)\\\\ &\text { When } x=5,\\\\ &y=f(5)=-3(\text { given })\\\\ &\text { Let } \left(x_{1}, y_{1}\right)=(5,-3)\\\\ \end{aligned}
The gradient of tangent to the curve at $x=5$ is $m=f^{\prime}(5)=4$ (given)
$\therefore$ The equation of tangent at $\left(x_{1}, y_{1}\right)$ is
\begin{aligned} &\\ y+3&=4(x-5)\\\\ 4 x-y&=23 \end{aligned}

23. If the tangent line to the curve $y=f(x)$ at $(4,3)$ cuts the $y$ - axis at $(0,2)$, find $f(4)$ and $f^{\prime}(4)$.

24. The tangent line to the curve $y=f(x)$ at $(4,3)$ cuts the $y$-axis at $(0,2)$.
\begin{aligned} &\\ \therefore\ f(4)&=3\\\\ \therefore\ f^{\prime}(4)&= \text{ gradient of tangent}\\\\ &=\frac{3-2}{4-0} \\\\ &=\frac{1}{4} \end{aligned}

25. If the tangent lines to the curve $y=4 x^{2}-x^{3}$ at the points where $x=-1$ and $x=2$ intersect at $P$, find the coordinates of the point $P$.

26. \begin{aligned} \text{Curve }: y&=4 x^{2}-x^{3}\\\\ \text{When } x&=-1,\\\\ y &=4(-1)^{2}-(-1)^{3} \\\\ &=5\\\\ \text{When } x&=2,\\\\ y &=4(2)^{2}-(2)^{3} \\\\ &=8\\\\ \end{aligned}
Let $\left(x_{1}, y_{1}\right)=(-1,5)$ and $\left(x_{2}, y_{2}\right)=(2,8)$.
\begin{aligned} &\\ \dfrac{d y}{d x}=8 x-3 x^{2}\\\\ \end{aligned}
Let the gradient of tangents at $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ be $m_{1}$ and $m_{2}$ respectively.
\begin{aligned} &\\ \therefore\ m_{1} &=\left.\dfrac{d y}{d x}\right|_{(-1,5)} \\\\ &=8(-1)-3(-1)^{2} \\\\ &=-11 \\\\ m_{2} &=\left.\dfrac{d y}{d x}\right|_{(2,8)} \\\\ &=8(2)-3(2)^{2} \\\\ &=4\\\\ \end{aligned}
Let the point of intersection of two tangents be $(a, b)$.
\begin{aligned} &\\ \dfrac{b-5}{a+1} &=-11 \\\\ 11 a+b &=-6 \\\\ \dfrac{b-8}{a-2} &=4 \\\\ 4 a-b &=0\\\\ \end{aligned}
Solving equations (1) and (2) yields $a=-\dfrac{2}{5}$ and $b=-\dfrac{8}{5}$
\begin{aligned} &\\ \therefore\ P=\left(-\dfrac{2}{5},-\dfrac{8}{5}\right) . \end{aligned}

27. Find the coordinates of point or points on the curve $y=x^{4}-2 x^{2}+3$ at which the curve has horizontal tangent(s).

28. \begin{aligned} \text{Curve }: y&=x^{4}-2 x^{2}+3\\\\ \frac{d y}{d x} &=4 x^{3}-4 x \\\\ &=4 x\left(x^{2}-1\right) \\\\ &=4 x(x+1)(x-1)\\\\ \end{aligned}
For horizontal tangents,
\begin{aligned} &\\ \frac{d y}{d x} &=0 . \\\\ 4 x(x+1)(x-1) &=0 \\\\ \therefore x=-1 \text { or } x &=0 \text { or } x=1 . \\\\ x=-1 \Rightarrow y &=(-1)^{4}-2(-1)^{2}+3=2 \\\\ x=0 \Rightarrow y &=(0)^{4}-2(0)^{2}+3=3 \\\\ x=1 \Rightarrow y &=(1)^{4}-2(1)^{2}+3=2\\\\ \end{aligned}
The points on the curve $y=x^{4}-2 x^{2}+3$ at which the curve has horizontal tangents are $(-1,2),(0,3)$ and $(1,2)$.

29. The curve $y=x^{3}+a x^{2}+b x+c$, where $a, b$ and $c$ are real constants, touches the $x$-axis at $x=1$ and cuts the $x$-axis at $x=4$. Find (i) the values of $a, b$ and $c$, (ii) the equation of the tangent to the curve at $x=0$.

30. Curve: $y=x^{3}+a x^{2}+b x+c\\\\$
Since the curve touches the $x$-axis at $x=1$ and cuts the $x$-axis at $x=4$, the points $(1,0)$ and $(4,0)$ lie on the curve and the gradient of tangent at $(1,0)$ is 0 .
\begin{aligned} &\\ \therefore(1)^{3}+a(1)^{2}+b(1)+c &=0 \\\\ a+b+c &=-1 \ldots(1)\\\\ (4)^{3}+a(4)^{2}+b(4)+c &=0 \\\\ 16 a+4 b+c &=-64 \ldots(2) \\\\ \text{By equation (2)- equation(1)} & \\\\ 15 a+3 b &=-63 \\\\ 5 a+b &=-21 \ldots(3)\\\\ \end{aligned}
The gradient of the tangent at any point $(x, y)$ on the curve is $\frac{d y}{d x}$.
\begin{aligned} &\\ \frac{d y}{d x} &=3 x^{2}+2 a x+b \\\\ \left.\frac{d y}{d x}\right|_{(1,0)} &=0 \\\\ 3+2 a+b &=0 \\\\ 2 a+b &=-3 \ldots(4)\\\\ \text{By equation (3)- equation(4)} & \\\\ 3 a &=-18 \\\\ a &=-6 \\\\ \text { Substituting } a &=-6 \text { in (4), } \\\\ -12+b &=-3 \\\\ b &=9 \\\\ \text { Substituting } a &=-6 \text { and } b=9 \text { in (1), } \\\\ -6+9+c &=-1 \\\\ c &=-4 \\\\ \therefore\ y &=x^{3}-6 x^{2}+9 x-4.\\\\ \end{aligned}
Substituting $a=-6$ and $b=9$ in (1),
When $x=0, y=-4\\\\$
Hence $(0,-4)$ lies on the curve.
$\therefore$ The gradient of the tangent at $(0,-4)$ on the curve is
\begin{aligned} &\\ \left.\frac{d y}{d x}\right|_{(0,-4)}& =b=9 \\\\ \therefore\ \text { The equation of tangent at } & (0,-4) \text { is } \\\\ y-(-4) &=9(x-0) \\\\ y &=9 x-4 . \end{aligned}

31. Find all values of $m$ such that the line $y=m x+3$ is tangent to the curve $y=\frac{1}{2} x^{2}-x+5$.

32. \begin{aligned} \text{ Curve }: y &= \frac{1}{2} x^{2}-x+5.\\\\ \text{ Gradient of tangent is } \frac{d y}{d x} &= x-1\\\\ \text{ Tangent }: y &= m x+3\\\\ \text{ Gradient of tangent } &= m\\\\ \therefore\ m &= x-1\\\\ x &= m+1\\\\ \text{Substituting } x &= m+1 \text{ in curve equation},\\\\ y &= \frac{1}{2}(m+1)^{2}-(m+1)+5\\\\ \text{Substituting } x &= m+1 \text{ in tangent equation}\\\\ y &= m(m+1)+3\\\\ \therefore\ \frac{1}{2}(m+1)^{2}-(m+1)+5 &= m(m+1)+3\\\\ m = -3 \text{ or } m &= 1 \end{aligned}

33. If the tangent to the curve $y=2 x^{3}-3 x+5$ at the point where $x=-1$ intersects the curve again at $A$, find the coordinates of $A$.

34. \begin{aligned} \text{ Curve }: y &= 2 x^{3}-3 x+5\\\\ \text{ When} x &= -1,\\\\ y &= 2(-1)^{3}-3(-1)+5 &= 6\\\\ \frac{d y}{d x} &= 6 x^{2}-3\\\\ &= 3\left(2 x^{2}-1\right)\\\\ m &= \left.\frac{d y}{d x}\right|_{(-1,6)}\\\\ &= 3\\\\ \end{aligned}
Let another point of intersection of curve and tangent be $(a, b)$.
\begin{aligned} &\\ \therefore b &= 2 a^{3}-3 a+5\\\\ \text{Since, } \frac{b-6}{a+1} &= 3\\\\ \frac{2 a^{3}-3 a+5-6}{a+1} &= 3\\\\ a &= 2\\\\ \therefore b &= 2(2)^{3}-3(2)+5\\\\ &= 15\\\\ \end{aligned}
Thus, the coordinates of the point $A$ is $(2,15)$.

35. If the lines passing through the point $(2,3)$ are tangent to the curve $y=3 x-x^{2}$, find the coordinates of the points on the curve where the tangents meet the curve.

36. \begin{aligned} &\text { Curve: } y=3 x-x^{2} \\\\ &\frac{d y}{d x}=3-2 x \\\\ &\text { Let }(a, b) \text { be the point on the curve } \\\\ &\text { where the tangent exists. } \\\\ &\therefore b=3 a-a^{2} \\\\ &m=\left.\frac{d y}{d x}\right|_{(a, b)}=3-2 a \\\\ &\text { Since the tangent at }(a, b) \text { pass } \\\\ &\text { through the point } (2,3), \\\\ &\frac{b-3}{a-2}=3-2 a \\\\ &\frac{3 a-a^{2}-3}{a-2}=3-2 a \\\\ &\therefore a=1(\text { or }) a=3 \\\\ &\text { When } a=1, b=3(1)-(1)^{2}=2 \\\\ &\text { When } a=3, b=3(3)-(3)^{2}=0 \\\\ &\therefore \text { The tangents meet the curve at }(1,2) \\\\ &\text { and }(3,0) \text {. } \end{aligned}

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