The tangent and normal to the ellipse 3 x ^ 2 + 5 y ^ 2 = 32 at the...

The equation of tangent and normal to the ellipse $\dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1$ at a point \left({x}_{1}, {y}_{1}\right)$are respectively$\dfrac{x{x}_{1}}{{a}^{2}} + \dfrac{y{y}_{1}}{{b}^{2}} = 1$and$\dfrac{{a}^{2}x}{{x}_{1}} - \dfrac{{b}^{2}y}{{y}_{1}} = {a}^{2} - {b}^{2} .$</p><p>The given ellipse is$3{x}^{2} + 5{y}^{2} = 32, ⇒ \dfrac{3{x}^{2}}{32} + \dfrac{5{y}^{2}}{32} = 1 .$</p><p>Hence, the tangent and normal to the ellipse$\dfrac{3{x}^{2}}{32} + \dfrac{5{y}^{2}}{32} = 1$at \left(2, 2\right)$ are respectively $\dfrac{3 \times 2 \times x}{32} + \dfrac{5 \times 2 \times y}{32} = 1$ and $\dfrac{32 \times x}{3 \times 2} - \dfrac{32 \times y}{5 \times 2} = \dfrac{32}{3} - \dfrac{32}{5}$

Hence, the tangent is $3x + 5y = 16 . . . \left(1\right)$

And, the normal is $5x - 3y = 4 . . . \left(2\right)$

The point where the tangent meets the $x$-axis can be obtained by putting $y = 0, ⇒ 3x + 0 = 16 ⇒ x = \dfrac{16}{3},$ thus the point $Q ≡ \left(\dfrac{16}{3}, 0\right) .$

Similarly, the point where the normal meets the $x$-axis, is $x = \dfrac{4}{5},$ thus the point $R ≡ \left(\dfrac{4}{3}, 0\right) .$

Tangent and normal intersect at $P\left(2, 2\right),$ and the length of perpendicular from any point on $x$-axis is the absolute value of its $y$-co-ordinate, hence the height of the triangle is $2$ units.

And, hence the required area is $= \dfrac{1}{2}\left(QR\right) \times h$

$= \dfrac{1}{2} \times \left(\dfrac{16}{3} - \dfrac{4}{5}\right) \times 2 = \dfrac{68}{15}$ sq units.