List of top Mathematics Questions on Tangents and Normals asked in KEAM

If \(x + 13y = 40\) is normal to the curve \(y = 5x^{2} + \alpha x + \beta\) at the point (1,3), then the value of \(\alpha\beta\) is equal to:
The equation of the normal to the curve given by $x^2 + 2x - 3y + 3 = 0$ at the point $(1,2)$ is
The equation of the tangent to the curve \( y = x + \frac{4}{x^2} \) that is parallel to the x-axis is
The equation of the tangent to the curve given by \( x^2 + 2x - 3y + 3 = 0 \) at the point \( (1,2) \) is
The equation of the tangent to the curve $y = x^3 - 6x + 5$ at $(2, 1)$ is:
Let $f(x) = 2x^3 - 5x^2 - 4x + 3,\ \frac{1}{2} \leq x \leq 3$. The point at which the tangent to the curve is parallel to the x-axis is:
The chord joining the points \( (5,5) \) and \( (11,227) \) on the curve \( y = 3x^2 - 11x - 15 \) is parallel to tangent at a point on the curve. The abscissa of the point is
The slope of the normal to the curve \( y = x^3 - 3x^2 + 2x + 1 \) at the point where the tangent is horizontal is
The slope of the tangent to the curve \( y = 3x^2 - 5x + 6 \) at \( (1,4) \) is
The slope of the normal to the curve \( y = x^2 - \frac{1}{x^2} \) at \( (-1, 0) \) is:
A straight line parallel to the line \( 2x - y + 5 = 0 \) is also a tangent to the curve \( y^2 = 4x + 5 \). Then the point of contact is:
The slope of the tangent to the curve \( y^2 e^{xy} = 9e^{-3}x^2 \) at \( (-1, 3) \) is:
The slope of the normal to the curve \( y = x^2 - \frac{1}{x^2} \) at \( (-1, 0) \) is:
A straight line parallel to the line \( 2x - y + 5 = 0 \) is also a tangent to the curve \( y^2 = 4x + 5 \). Then the point of contact is:
The slope of the tangent to the curve \( y^2 e^{xy} = 9e^{-3}x^2 \) at \( (-1, 3) \) is:
The slope of the normal to the curve \( y = x^2 - \frac{1}{x^2} \) at \( (-1, 0) \) is:
The slope of the tangent to the curve \( y^2 e^{xy} = 9e^{-3}x^2 \) at \( (-1, 3) \) is:
A straight line parallel to the line \( 2x - y + 5 = 0 \) is also a tangent to the curve \( y^2 = 4x + 5 \). Then the point of contact is: