Let the tangents to the curve x2+2x−4y+9=0 at the point P(1,3) on it meet the y-axis at A. Let the line passing through P and parallel to the line x−3y=6 meet the parabola y2=4x at B. If B lies on the line 2x−3y=8, then (AB)2 is equal to _______
Explanation
C:x2+2x−4y+9=0
C:(x+1)2=4(y−2)
Tangent at P(1,3)
xx1+yy1+g(x+x1)+f(y+y1)+C=0
⇒x⋅1+(x+1)−2(y+3)+9=0
⇒x−y+2=0
A:(0,2)
Line is parallel to x−3y=6 passes through (1,3) is x−3y+8=0
Meet parabola y2=4x
⇒y2=4(3y−8)
⇒(y−8)(y−4)=0
Point of intersection are (4,4) and (16,8). (16,8) lies on 2x−3y=8
Hence A(0,2)
B(16,8)
(AB)2=256+36=292
Explanation
C:x2+2x−4y+9=0
C:(x+1)2=4(y−2)
Tangent at P(1,3)
xx1+yy1+g(x+x1)+f(y+y1)+C=0
⇒x⋅1+(x+1)−2(y+3)+9=0
⇒x−y+2=0
A:(0,2)
Line is parallel to x−3y=6 passes through (1,3) is x−3y+8=0
Meet parabola y2=4x
⇒y2=4(3y−8)
⇒(y−8)(y−4)=0
Point of intersection are (4,4) and (16,8). (16,8) lies on 2x−3y=8
Hence A(0,2)
B(16,8)
(AB)2=256+36=292
