Question

Find equation of tangent and the normal x^2+y^2-3x+10y-15=0 at (4, -11)

Answer 1

Step-by-step explanation:

Given, equation of circle ${x}^{2}+{y}^{2}-3x+10y-15=0$

Differentiating both sides w.r.t x,

$2x + 2y \frac{dy}{dx} - 3 + 10 \frac{dy}{dx} = 0$

$\implies \: (2x - 3)+ (2y + 10)\frac{dy}{dx} = 0$

$\implies \:\frac{dy}{dx} = \frac{3 - 2x}{10 + 2y}$

Now,

$\implies \:m = \frac{dy}{dx} \bigg | _{(4 , - 11)} = \frac{3 - 2(4)}{10 + 2( - 11)} = \frac{ - 5}{ - 12} = \frac{5}{12}$

$\large\tt{\bold{Equation\:\: of \:\:tangent:}}$

$(y + 11) = m(x - 4)$

$\implies \: y + 11 = \frac{5}{12} (x - 4)$

$\implies \: 12y + 132 = 5 (x - 4)$

$\implies \: 12y + 132 = 5x - 20$

$\implies \: 5x - 12y-152 = 0$

$\large\tt{\bold{Equation \:\:of \:\:normal:}}$

$(y + 11)m + (x - 4) = 0$

$\implies \: (y + 11) \frac{5}{12} + (x - 4) = 0$

$\implies \: (y + 11) 5+ 12(x - 4) = 0$

$\implies \: 5y + 55+ 12x - 48 = 0$

$\implies \: 12x + 5y + 55 - 48 = 0$

$\implies \: 12x + 5y +7 = 0$