Advertisements
Advertisements
प्रश्न
The equation of tangent to the curve y = 3x2 - x + 1 at the point (1, 3) is
(a) y=5x+2
(b)y=5x-2
(c)y=1/5x+2
(d)y=1/5x-2
उत्तर
y = 5x - 2
`dy/dx=6x-1" at "(1,3)`
Slope of the tangent at (1, 3) = (6 - 1) = 5
Equation of tangent is y - y1 = m(x - x1)
y - 3 = 5(x - 1)
5x - y - 2 = 0
y = 5x - 2
APPEARS IN
संबंधित प्रश्न
The equation of tangent to the curve y=`y=x^2+4x+1` at
(-1,-2) is...............
(a) 2x -y = 0 (b) 2x+y-5 = 0
(c) 2x-y-1=0 (d) x+y-1=0
If the line y = x + k touches the hyperbola 9x2 -16y2 = 144, then k = _________
Show that the line x+ 2y + 8 = 0 is tangent to the parabola y2 = 8x. Hence find the point of contact
Show that the product of lengths of perpendicular segments drawn from the foci to any tangent to the hyperbola `x^2/25 + y^2/16 = 1` is equal to 16.
Find the equation of tangent to the curve y = 3x2 − x + 1 at P(1, 3).
