Question

At any point ( x , y ) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (–4, –3). Find the equation of the curve given that it passes through (–2, 1).

Answer 1

Since it is given that the point ( x,y )is the point of contact on the curve to its tangent. Slope of the line segment which joins ( x,y ) and ( −4,−3 ) is given as,

m 1 = y 2 − y 1 x 2 − x 1 m 1 = y−( −3 ) x−( −4 ) m 1 = y+3 x+4

Also slope of the tangent is given as dy dx then,

dy dx = m 2

According to the question, slope of tangent is twice the slope of line segment. So,

m 2 =2 m 1 dy dx = 2( y+3 ) ( x+4 ) dy ( y+3 ) = 2dx ( x+4 )

Integrate both sides.

∫ dy ( y+3 ) = ∫ 2dx ( x+4 ) log( y+3 )=2log( x+4 )+logC log( y+3 )=log( C ( x+4 ) 2 ) y+3=C ( x+4 ) 2 (1)

Above equation is the required general equation of the curve that passes through point ( −2,1 ),

1+3=C ( −2+4 ) 2 C=1

Substitute the values into equation (1),

y+3= ( x+4 ) 2

Thus, the required equation of the curve that passes through point ( −2,1 ) is y+3= ( x+4 ) 2 .