Question

find the equatiom of the tangent to the ellipse 2x^2+3y^2=5 which is the perpendicular to the line 3x+2y+7=0​

Answer 1

let us first bring the equation in standard form

x²/5/2+y²/5/3=1

a²=5/2

b²=5/3

since its a tangent therefore its general equation is

xx₁/a²+yy₁/b²=1

where x1=acosx

y1=bsinx

so eq is

xcosx/a+ysinx/b=1

y=1-xcosx/a*b/sinx

y=1-cotx*b/a*x-----1

as we know that slope of the tangent is

m=-1/m1

where m1 is slope of 3x+2y+7

m=2/3

so y=2x/3+2c/3-----2

1=2

now just compare the two equations to find c

and substitute in eq 2 to get the answer

Answer 2

Heya!!!

LET THE TANGENT BE DRAWN TO THE CURVE AT POINT ( x1 , y1 )

=>

EQUATION OF TANGENT TO THE ELLIPSE IS

2 ( x × x1 ) + 3 ( y × y1 ) = 5

Slope of ellipse ( m1 ) = -2x1 /3y1

And

Slope of given line is ( m2 ) = -3/2

(m1 × m2 ) = -1 ( y?)

becoz product of slopes = -1 when two lines are perpendicular.

=>

x1 = ±1 And y1 =±1

So, Equation of tangent to the ellipse is

2 ( x × 1) + 3 ( y × 1 ) = 5

= 2x +3y = 5


Or


2 ( x × -1 ) + 3 ( y × -1 ) = 5



2x + 3y = -5.