Question

ANSWER FOR 45 POINT


determine the point on the curve x^2+y^2=13 where the tangent are perpendicular to the line 3x-2y=0 ​

Answer 1

$\huge\underline\mathbb\red{GIVEN-:}$

$the \: given \: curve \: is \: \:x ^{2} + {y}^{2} = 13 .......(1)$

$let (x1,y1) be a point on (1)$

$\huge\underline\mathbb\blue{THEN}$

${x}1^{2} + {y}^{2} = 13.........(2)$

diff.(1) W.R.T X, 2X + 2Y dy/dx =0

⇒ dy/dx = -x/y ⇒ dy/dx(x1 , y1) =-x1/y1.

$but \: the \: tangents \: ar e \: perendicular \: to \: 3x - 2y = 0$

$∴( \frac{ - x1}{y1} )( - \frac{3}{ - 2}) = - 1 \: \: \: \: \: \: \: (m1.m2 = 1)$

⇒

$\frac{3x1}{2y1} ⇒3x1 \: = 2y1.......(3)$

$from \: (3) \: y1 = \frac{3x1}{2} .......(4)$

putting (2)

${x}1^{2} + \frac{9x1}{4} = 13$

⇒

$13 {x}1^{2} \div 4 = 13 \: ⇒ {x}1^{2} = 4 \: ⇒x1 = + - 2$

$\huge\underline\mathbb\green{WHEN}$

x1=2

then from (4), y1 = 3/2 ×2 =3.

$\huge\underline\mathbb\pink{WHEN}$

x1= -2 , the from (4) y2 = 3/2 ×-2 =-3

∴REQUIRED POINTS ARE (2,3) & (-2,-3)✅