Question

A chord is a normal to a parabola and is inclined at an angle to the axis; prove that the area of the triangle formed by it and the tangents it its extermites is $4a^{2}se{c}^3\theta cose{c}^3\theta$

Answer 1

Let the extremities of the normal chord be P and Q and the tangents at P and Q to the parabola say $y^2=4ax$ meet in T. Let the co-ordinates of T be $(x_1,y_2)$.
PQ will be the chord of contact for T with respect to teh parabola so area of triangle TPQ will be
$[y_1^2-4a{x}_1{]}^{\frac{3}{2}}2a$ ....(1)
The equation to the chord of contact of T will be
$y{y}_1=2a(x+x_1)$
$\Rightarrow y=\frac{2ax}{y_1}+\frac{2a{x}_1}{y_1}$ ... (2)
Equation to any normal to y^2=4ax is
$y=mx-2am-a{m}^2$ ......(3)
So (2) and (3) must be identical. As the coefficients of y are equal, others must also equal, so
$m=\frac{2a}{y_1}and-2am-a{m}^3=\frac{2a{x}_1}{y_1}$
hence $y_1=\frac{2a}{m}and{x}_1=(-2a-a{m}^2)$.
If the inclination of the chord of contact, i.e., the normal is \theta; then
$m=tan\theta$.
So $y_1=\frac{2a}{m}=2acot\theta and{x}_1(-2a-a{m}^2\theta )$.
Substituting in (1), we get
Area of triangle =$[4a^{2}co{t}^2\theta -4a(-2a-ata{n}^2\theta ){]}^{\frac{\frac{3}{2}}{2a}}$
$=\frac{1}{2a}[4a^{2}co{t}^2\theta +8a^{2}4a^{2}ta{n}^2\theta {]}^{\frac{3}{2}}=\frac{1}{2a}[4a^2(co{t}^2\theta +2ta{n}^2\theta ){]}^{\frac{3}{2}}$
$=\frac{1}{2a}(4a^2{)}^{\frac{3}{2}}{\left\{\right(cot\theta +tan\theta \left)\right\}}^{\frac{3}{2}}$
$=\frac{1}{2a}8a^3{\left\{\right(sec\theta +cosec\theta {)}^2\}}^{\frac{3}{2}}$
$[\because cot\theta +tan\theta =\frac{co{s}^2\theta +si{n}^2\theta }{sin\theta .cos\theta }=sec\theta .cosec\theta ]$
$=4a^{2}se{c}^3\theta .cose{c}^3\theta$.