Question

1. Find the equations of the tangent and normal to the circle



(i) 2x2+2y2 + 3x- 4y + 1= 0 at (-1, 2)​

Answer 1

$\textbf{Given:}$

$\textsf{Circle is}$

$\mathsf{2x^2+2y^2+3x-4y+1=0}$

$\textbf{To find:}$

$\textsf{Equation of tangent and normal at (-1,2)}$

$\textbf{Solution:}$

$\textsf{The equation of tangent will be of the form}$

$\mathsf{2x\,x_1+2y\,y_1+3\left(\dfrac{x+x_1}{2}\right)-4\left(\dfrac{y+y_1}{2}\right)+1=0}$

$\mathsf{Here,\;(x_1,y_1)=(-1,2)}$

$\mathsf{2x(-1)+2y(2)+3\left(\dfrac{x-1}{2}\right)-4\left(\dfrac{y+2}{2}\right)+1=0}$

$\mathsf{-2x+4y+3\left(\dfrac{x-1}{2}\right)-4\left(\dfrac{y+2}{2}\right)+1=0}$

$\mathsf{-4x+8y+3(x-1)-4(y+2)+2=0}$

$\mathsf{-4x+8y+3x-3-4y-8+2=0}$

$\mathsf{-x+4y-9=0}$

$\textsf{The equation of tangent is}$

$\implies\boxed{\mathsf{x-4y+9}=0}$

$\textsf{Equation of normal can be written as}$

$\mathsf{-4x-y+k=0}$

$\textsf{But this passes through (-1,2)}$

$\mathsf{-4(-1)-2+k=0}$

$\mathsf{4-2+k=0}$

$\mathsf{2+k=0}$

$\mathsf{k=-2}$

$\therefore\textsf{Equation of normal is}$

$\mathsf{-4x-y-2=0}$

$\implies\boxed{\mathsf{4x+y+2=0}}$

$\textbf{Find more:}$

Let a be the point where the curve 5a^2x^3+10ax^2+x+2y-4 meeys y-axis. then the equation of tangent to the curve at the point a meets the curve again is?

https://brainly.in/question/16627156