1.07m Tangents and normals: gradient and equations

1. The curve \(C\) has equation \(y = \mathrm { f } ( x )\) where

$$f ( x ) = \frac { 4 x + 1 } { x - 2 } , \quad x > 2$$

1. Show that $$f ^ { \prime } ( x ) = \frac { - 9 } { ( x - 2 ) ^ { 2 } }$$ Given that \(P\) is a point on \(C\) such that \(\mathrm { f } ^ { \prime } ( x ) = - 1\),
2. find the coordinates of \(P\).
   

3. The curve \(C\) has equation \(x = 8 y \tan 2 y\) The point \(P\) has coordinates \(\left( \pi , \frac { \pi } { 8 } \right)\)

1. Verify that \(P\) lies on \(C\).
2. Find the equation of the tangent to \(C\) at \(P\) in the form \(a y = x + b\), where the constants \(a\) and \(b\) are to be found in terms of \(\pi\).

5. The point \(P\) lies on the curve with equation $$x = ( 4 y - \sin 2 y ) ^ { 2 }$$ Given that \(P\) has \(( x , y )\) coordinates \(\left( p , \frac { \pi } { 2 } \right)\), where \(p\) is a constant,

1. find the exact value of \(p\). The tangent to the curve at \(P\) cuts the \(y\)-axis at the point \(A\).
2. Use calculus to find the coordinates of \(A\).

6. $$f ( x ) = \frac { x ^ { 4 } + x ^ { 3 } - 3 x ^ { 2 } + 7 x - 6 } { x ^ { 2 } + x - 6 } , \quad x > 2 , x \in \mathbb { R }$$

1. Given that $$\frac { x ^ { 4 } + x ^ { 3 } - 3 x ^ { 2 } + 7 x - 6 } { x ^ { 2 } + x - 6 } \equiv x ^ { 2 } + A + \frac { B } { x - 2 }$$ find the values of the constants \(A\) and \(B\).
2. Hence or otherwise, using calculus, find an equation of the normal to the curve with equation \(y = \mathrm { f } ( x )\) at the point where \(x = 3\)

5. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve \(C\) with equation $$y = 2 \ln ( 2 x + 5 ) - \frac { 3 x } { 2 } , \quad x > - 2.5$$ The point \(P\) with \(x\) coordinate - 2 lies on \(C\).

1. Find an equation of the normal to \(C\) at \(P\). Write your answer in the form \(a x + b y = c\), where \(a\), \(b\) and \(c\) are integers. The normal to \(C\) at \(P\) cuts the curve again at the point \(Q\), as shown in Figure 2
2. Show that the \(x\) coordinate of \(Q\) is a solution of the equation $$x = \frac { 20 } { 11 } \ln ( 2 x + 5 ) - 2$$ The iteration formula $$x _ { n + 1 } = \frac { 20 } { 11 } \ln \left( 2 x _ { n } + 5 \right) - 2$$ can be used to find an approximation for the \(x\) coordinate of \(Q\).
3. Taking \(x _ { 1 } = 2\), find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving each answer to 4 decimal places.

4. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve \(C\) with equation $$y = \mathrm { e } ^ { - 2 x } + x ^ { 2 } - 3$$ The curve \(C\) crosses the \(y\)-axis at the point \(A\). The line \(l\) is the normal to \(C\) at the point \(A\).

1. Find the equation of \(l\), writing your answer in the form \(y = m x + c\), where \(m\) and \(c\) are constants. The line \(l\) meets \(C\) again at the point \(B\), as shown in Figure 1 .
2. Show that the \(x\) coordinate of \(B\) is a solution of $$x = \sqrt { 1 + \frac { 1 } { 2 } x - \mathrm { e } ^ { - 2 x } }$$ Using the iterative formula $$x _ { n + 1 } = \sqrt { 1 + \frac { 1 } { 2 } x _ { n } - \mathrm { e } ^ { - 2 x _ { n } } }$$ with \(x _ { 1 } = 1\)
3. find \(x _ { 2 }\) and \(x _ { 3 }\) to 3 decimal places.

7.

1. Sketch the curve with equation \(y = \ln x\).
2. Show that the tangent to the curve with equation \(y = \ln x\) at the point ( \(\mathrm { e } , 1\) ) passes through the origin.
3. Use your sketch to explain why the line \(y = m x\) cuts the curve \(y = \ln x\) between \(x = 1\) and \(x = \mathrm { e }\) if \(0 < m < \frac { 1 } { \mathrm { e } }\). Taking \(x _ { 0 } = 1.86\) and using the iteration \(x _ { n + 1 } = \mathrm { e } ^ { \frac { 1 } { 3 } x _ { n } }\),
4. calculate \(x _ { 1 } , x _ { 2 } , x _ { 3 } , x _ { 4 }\) and \(x _ { 5 }\), giving your answer to \(x _ { 5 }\) to 3 decimal places. The root of \(\ln x - \frac { 1 } { 3 } x = 0\) is \(\alpha\).
5. By considering the change of sign of \(\ln x - \frac { 1 } { 3 } x\) over a suitable interval, show that your answer for \(x _ { 5 }\) is an accurate estimate of \(\alpha\), correct to 3 decimal places.

   7. continued

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6. A curve has equation $$4 y ^ { 2 } + 3 x = 6 y \mathrm { e } ^ { - 2 x }$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\). The curve crosses the \(y\)-axis at the origin and at the point \(P\).
2. Find the equation of the normal to the curve at \(P\), writing your answer in the form \(y = m x + c\) where \(m\) and \(c\) are constants to be found.

8. \begin{figure}[h] \end{figure} \section*{In this question you must show all stages of your working.} \section*{Solutions relying entirely on calculator technology are not acceptable.} A curve \(C\) has parametric equations $$x = \sin ^ { 2 } t \quad y = 2 \tan t \quad 0 \leqslant t < \frac { \pi } { 2 }$$ The point \(P\) with parameter \(t = \frac { \pi } { 4 }\) lies on \(C\).
The line \(l\) is the normal to \(C\) at \(P\), as shown in Figure 3.

1. Show, using calculus, that an equation for \(l\) is $$8 y + 2 x = 17$$ The region \(S\), shown shaded in Figure 3, is bounded by \(C , l\) and the \(x\)-axis.
2. Find, using calculus, the exact area of \(S\).

9. \begin{figure}[h] \end{figure} Figure 4 shows a sketch of the curve \(C\) with parametric equations $$x = \sec t \quad y = \sqrt { 3 } \tan \left( t + \frac { \pi } { 3 } \right) \quad \frac { \pi } { 6 } < t < \frac { \pi } { 2 }$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\)
2. Find an equation for the tangent to \(C\) at the point where \(t = \frac { \pi } { 3 }\) Give your answer in the form \(y = m x + c\), where \(m\) and \(c\) are constants.
3. Show that all points on \(C\) satisfy the equation $$y = \frac { A x ^ { 2 } + B \sqrt { 3 x ^ { 2 } - 3 } } { 4 - 3 x ^ { 2 } }$$ where \(A\) and \(B\) are constants to be found.

1. A curve has parametric equations

$$x = 7 \cos t - \cos 7 t , y = 7 \sin t - \sin 7 t , \quad \frac { \pi } { 8 } < t < \frac { \pi } { 3 }$$

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\). You need not simplify your answer.
2. Find an equation of the normal to the curve at the point where \(t = \frac { \pi } { 6 }\). Give your answer in its simplest exact form.
   

1. The curve \(C\) has parametric equations

$$x = \ln t , \quad y = t ^ { 2 } - 2 , \quad t > 0$$ Find

1. an equation of the normal to \(C\) at the point where \(t = 3\),
2. a cartesian equation of \(C\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{a3ece8a8-8107-4c3a-a6a9-c19b5e35ec5a-10_579_759_740_571} \captionsetup{labelformat=empty} \caption{Figure 1}
   \end{figure} The finite area \(R\), shown in Figure 1, is bounded by \(C\), the \(x\)-axis, the line \(x = \ln 2\) and the line \(x = \ln 4\). The area \(R\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.
3. Use calculus to find the exact volume of the solid generated.

1. The curve \(C\) has the equation \(2 x + 3 y ^ { 2 } + 3 x ^ { 2 } y = 4 x ^ { 2 }\).

The point \(P\) on the curve has coordinates \(( - 1,1 )\).

1. Find the gradient of the curve at \(P\).
2. Hence find the equation of the normal to \(C\) at \(P\), giving your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.
   

7. The curve \(C\) has parametric equations $$x = 2 \cos t , \quad y = \sqrt { 3 } \cos 2 t , \quad 0 \leqslant t \leqslant \pi$$ where \(t\) is a parameter.

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\). The point \(P\) lies on \(C\) where \(t = \frac { 2 \pi } { 3 }\) The line \(l\) is a normal to \(C\) at \(P\).
2. Show that an equation for \(l\) is $$2 x - 2 \sqrt { 3 } y - 1 = 0$$ The line \(l\) intersects the curve \(C\) again at the point \(Q\).
3. Find the exact coordinates of \(Q\). You must show clearly how you obtained your answers. \includegraphics[max width=\textwidth, alt={}, center]{245bbe52-3a14-4494-af17-7711caf79b22-23_106_63_2595_1882}

1. A curve has parametric equations

$$x = 2 \cot t , \quad y = 2 \sin ^ { 2 } t , \quad 0 < t \leqslant \frac { \pi } { 2 }$$

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of the parameter \(t\).
2. Find an equation of the tangent to the curve at the point where \(t = \frac { \pi } { 4 }\).
3. Find a cartesian equation of the curve in the form \(y = \mathrm { f } ( x )\). State the domain on which the curve is defined.
   

1. A curve \(C\) is described by the equation

$$3 x ^ { 2 } - 2 y ^ { 2 } + 2 x - 3 y + 5 = 0$$ Find an equation of the normal to \(C\) at the point ( 0,1 ), giving your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.

4. \begin{figure}[h] \end{figure} The curve shown in Figure 2 has parametric equations $$x = \sin t , y = \sin \left( t + \frac { \pi } { 6 } \right) , \quad - \frac { \pi } { 2 } < t < \frac { \pi } { 2 }$$

1. Find an equation of the tangent to the curve at the point where \(t = \frac { \pi } { 6 }\).
2. Show that a cartesian equation of the curve is $$y = \frac { \sqrt { } 3 } { 2 } x + \frac { 1 } { 2 } \sqrt { } \left( 1 - x ^ { 2 } \right) , \quad - 1 < x < 1$$

6. A curve has parametric equations $$x = \tan ^ { 2 } t , \quad y = \sin t , \quad 0 < t < \frac { \pi } { 2 }$$

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\). You need not simplify your answer.
2. Find an equation of the tangent to the curve at the point where \(t = \frac { \pi } { 4 }\). Give your answer in the form \(y = a x + b\), where \(a\) and \(b\) are constants to be determined.
3. Find a cartesian equation of the curve in the form \(y ^ { 2 } = \mathrm { f } ( x )\).
   \section*{LU}

4. \begin{figure}[h] \end{figure} Figure 1 shows part of the curve with parametric equations $$x = \tan t , \quad y = \sin 2 t , \quad - \frac { \pi } { 2 } < t < \frac { \pi } { 2 } .$$

1. Find the gradient of the curve at the point \(P\) where \(t = \frac { \pi } { 3 }\).
2. Find an equation of the normal to the curve at \(P\).
3. Find an equation of the normal to the curve at the point \(Q\) where \(t = \frac { \pi } { 4 }\).

8. The parabola \(C\) has equation \(y ^ { 2 } = 4 a x\), where \(a\) is a positive constant. The point \(P \left( a p ^ { 2 } , 2 a p \right)\) lies on the parabola \(C\).

1. Show that an equation of the tangent to \(C\) at \(P\) is $$p y = x + a p ^ { 2 }$$ The tangent to \(C\) at the point \(P\) intersects the directrix of \(C\) at the point \(B\) and intersects the \(x\)-axis at the point \(D\). Given that the \(y\)-coordinate of \(B\) is \(\frac { 5 } { 6 } a\) and \(p > 0\),
2. find, in terms of \(a\), the \(x\)-coordinate of \(D\). Given that \(O\) is the origin,
3. find, in terms of \(a\), the area of the triangle \(O P D\), giving your answer in its simplest form.

4. The parabola \(C\) has cartesian equation \(y ^ { 2 } = 12 x\) The point \(P \left( 3 p ^ { 2 } , 6 p \right)\) lies on \(C\), where \(p \neq 0\)

1. Show that the equation of the normal to the curve \(C\) at the point \(P\) is $$y + p x = 6 p + 3 p ^ { 3 }$$ This normal crosses the curve \(C\) again at the point \(Q\).
   Given that \(p = 2\) and that \(S\) is the focus of the parabola, find
2. the coordinates of the point \(Q\),
3. the area of the triangle \(P Q S\).

6. The rectangular hyperbola \(H\) has equation \(x y = c ^ { 2 }\), where \(c\) is a non-zero constant. The point \(P \left( c p , \frac { c } { p } \right)\), where \(p \neq 0\), lies on \(H\).

1. Show that the normal to \(H\) at \(P\) has equation $$y p - p ^ { 3 } x = c \left( 1 - p ^ { 4 } \right)$$ The normal to \(H\) at \(P\) meets \(H\) again at the point \(Q\).
2. Find, in terms of \(c\) and \(p\), the coordinates of \(Q\).

8. The parabola \(C\) has equation \(y ^ { 2 } = 4 a x\), where \(a\) is a positive constant. The point \(P \left( a t ^ { 2 } , 2 a t \right)\) lies on \(C\).

1. Using calculus, show that the normal to \(C\) at \(P\) has equation $$y + t x = a t ^ { 3 } + 2 a t$$ The point \(S\) is the focus of the parabola \(C\).
   The point \(B\) lies on the positive \(x\)-axis and \(O B = 5 O S\), where \(O\) is the origin.
2. Write down, in terms of \(a\), the coordinates of the point \(B\). A circle has centre \(B\) and touches the parabola \(C\) at two distinct points \(Q\) and \(R\). Given that \(t \neq 0\),
3. find the coordinates of the points \(Q\) and \(R\).
4. Hence find, in terms of \(a\), the area of triangle \(B Q R\).

1. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

The rectangular hyperbola \(H\) has equation \(x y = 20\) The point \(P \left( 2 t \sqrt { a } , \frac { 2 \sqrt { a } } { t } \right) , t \neq 0\), where \(a\) is a constant, is a general point on \(H\)

1. State the value of \(a\)
2. Show that the normal to \(H\) at the point \(P\) has equation $$t y - t ^ { 3 } x - 2 \sqrt { 5 } \left( 1 - t ^ { 4 } \right) = 0$$ The points \(A\) and \(B\) lie on \(H\) The point \(A\) has parameter \(t = c\) and the point \(B\) has parameter \(t = - \frac { 1 } { 2 c }\), where \(c\) is a constant. The normal to \(H\) at \(A\) meets \(H\) again at \(B\)
3. Determine the possible values of \(C\)

1. The hyperbola \(H\) has equation \(x y = c ^ { 2 }\) where \(c\) is a positive constant.

The point \(P \left( c t , \frac { c } { t } \right)\), where \(t > 0\), lies on \(H\).
The tangent to \(H\) at \(P\) meets the \(x\)-axis at the point \(A\) and meets the \(y\)-axis at the point \(B\).

1. Determine, in terms of \(c\) and \(t\),
   1. the coordinates of \(A\),
   2. the coordinates of \(B\). Given that the area of triangle \(A O B\), where \(O\) is the origin, is 90 square units,
2. determine the value of \(c\), giving your answer as a simplified surd.