Ellipse tangent/normal equation derivation

9. The ellipse \(E\) has equation $$\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 16 } = 1$$ The point \(P\) lies on the ellipse and has coordinates \(( 5 \cos \theta , 4 \sin \theta )\) where \(0 < \theta < \frac { \pi } { 2 }\) The line \(l\) is the normal to the ellipse at the point \(P\).

1. Show that an equation for \(l\) is $$5 x \sin \theta - 4 y \cos \theta = 9 \sin \theta \cos \theta$$ The point \(F\) is the focus of \(E\) that lies on the positive \(x\)-axis.
2. Determine the coordinates of \(F\). The line \(l\) crosses the \(x\)-axis at the point \(Q\).
3. Show that $$\frac { | Q F | } { | P F | } = e$$ where \(e\) is the eccentricity of \(E\).
   

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2. An ellipse has equation $$\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 4 } = 1$$ The point \(P\) lies on the ellipse and has coordinates \(( 5 \cos \theta , 2 \sin \theta ) , 0 < \theta < \frac { \pi } { 2 }\) The line \(L\) is a normal to the ellipse at the point \(P\).

1. Show that an equation for \(L\) is $$5 x \sin \theta - 2 y \cos \theta = 21 \sin \theta \cos \theta$$ Given that the line \(L\) crosses the \(y\)-axis at the point \(Q\) and that \(M\) is the midpoint of \(P Q\),
2. find the exact area of triangle \(O P M\), where \(O\) is the origin, giving your answer as a multiple of \(\sin 2 \theta\) WIHN SIHI NITIIUM ION OC
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1. The ellipse \(E\) has equation

$$\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 9 } = 1$$ The line \(l\) is the normal to \(E\) at the point \(P ( 5 \cos \theta , 3 \sin \theta )\) where \(0 < \theta < \frac { \pi } { 2 }\)

1. Using calculus, show that an equation for \(l\) is $$5 x \sin \theta - 3 y \cos \theta = 16 \sin \theta \cos \theta$$ Given that
   • \(\quad l\) intersects the \(y\)-axis at the point \(Q\)
   • the midpoint of the line segment \(P Q\) is \(M\)
   • determine the exact maximum area of triangle \(O M P\) as \(\theta\) varies, where \(O\) is the origin.
   You must justify your answer.

1. The ellipse \(E\) has equation

$$\frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1 , \quad a > b > 0$$ The line \(l\) is a normal to \(E\) at a point \(P ( a \cos \theta , b \sin \theta ) , \quad 0 < \theta < \frac { \pi } { 2 }\)

1. Using calculus, show that an equation for \(l\) is $$a x \sin \theta - b y \cos \theta = \left( a ^ { 2 } - b ^ { 2 } \right) \sin \theta \cos \theta$$ The line \(l\) meets the \(x\)-axis at \(A\) and the \(y\)-axis at \(B\).
2. Show that the area of the triangle \(O A B\), where \(O\) is the origin, may be written as \(k \sin 2 \theta\), giving the value of the constant \(k\) in terms of \(a\) and \(b\).
3. Find, in terms of \(a\) and \(b\), the exact coordinates of the point \(P\), for which the area of the triangle \(O A B\) is a maximum.

5. The ellipse \(E\) has equation $$x ^ { 2 } + 9 y ^ { 2 } = 9$$ The point \(P ( a \cos \theta , b \sin \theta )\) is a general point on the ellipse \(E\).

1. Write down the value of \(a\) and the value of \(b\). The line \(L\) is a tangent to \(E\) at the point \(P\).
2. Show that an equation of the line \(L\) is given by $$3 y \sin \theta + x \cos \theta = 3$$ The line \(L\) meets the \(x\)-axis at the point \(Q\) and meets the \(y\)-axis at the point \(R\).
3. Show that the area of the triangle \(O Q R\), where \(O\) is the origin, is given by $$k \operatorname { cosec } 2 \theta$$ where \(k\) is a constant to be found. The point \(M\) is the midpoint of \(Q R\).
4. Find a cartesian equation of the locus of \(M\), giving your answer in the form \(y ^ { 2 } = \mathrm { f } ( x )\).

2. The ellipse \(E\) has equation $$\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 25 } = 1$$ The line \(l\) is the normal to \(E\) at the point \(P ( 6 \cos \theta , 5 \sin \theta )\), where \(0 < \theta < \frac { \pi } { 2 }\)

1. Use calculus to show that an equation of \(l\) is $$6 x \sin \theta - 5 y \cos \theta = 11 \sin \theta \cos \theta$$ The line \(l\) meets the \(x\)-axis at the point \(Q\). The point \(R\) is the foot of the perpendicular from \(P\) to the \(x\)-axis.
2. Show that \(\frac { O Q } { O R } = e ^ { 2 }\), where \(e\) is the eccentricity of the ellipse \(E\).

An ellipse has equation $$\frac{x^2}{25} + \frac{y^2}{4} = 1$$ The point \(P\) lies on the ellipse and has coordinates \((5\cos \theta, 2\sin \theta)\), \(0 < \theta < \frac{\pi}{2}\) The line \(L\) is a normal to the ellipse at the point \(P\).

1. Show that an equation for \(L\) is $$5x \sin \theta - 2y \cos \theta = 21 \sin \theta \cos \theta$$ [5]

Given that the line \(L\) crosses the \(y\)-axis at the point \(Q\) and that \(M\) is the midpoint of \(PQ\),

1. find the exact area of triangle \(OPM\), where \(O\) is the origin, giving your answer as a multiple of \(\sin 2\theta\) [6]