Find the equations of the tangent and normal to the curve 16x^2 + 9y^…

The correct answer is Option (2) → Tangent: $32x+27y−145=0$; Normal: $27x−32y+42=0$

The given curve is $16x^2 +9y^2 = 145$   ...(i)

Since $(x_1,y_1)$ where $x_1 = 2, y_1> 0$ lies on curve (i), we get

$16.2^2+9{y_1}^2 = 145 ⇒9{y_1}^2 = 145-64⇒ 9{y_1}^2 = 81$

$⇒{y_1}^2=9⇒y_1 = 3, -3$ but $y_1>0⇒y_1 = 3$.

Therefore, the point $(x_1,y_1)$ is (2, 3).

Diff. (i) w.r.t. x, we get $16.2x + 9.2y\frac{dy}{dx}=0⇒\frac{dy}{dx}=-\frac{16x}{9y}$,

∴ the slope of the tangent to the curve (i) at the point (2, 3) = $-\frac{16.2}{9.3}=-\frac{32}{27}$

∴ The equation of the tangent to the given curve at (2, 3) is

$y-3=-\frac{32}{27} (x-2)$ or $32x + 27y -145=0$.

The slope of the normal to the given curve at the point (2, 3) = $\frac{27}{32}$   $(m_2=-\frac{1}{m_1})$

∴ The equation of the normal to the given curve at (2, 3) is

$y-3=\frac{27}{32}(x-2)$ or $27x-32y+42 = 0$.