Diameter Form of Circle Equation The equation of the circle drawn on the straight line joining two points \((x_1, y_1)\) and \((x_2, y_2)\) is (\(x-x_1\))(\(x-x_2\)) + (\(y-y_1\))(\(y-y_2\)) = 0. This is known as diameter form of circle. where \((x_1, y_1)\) and \((x_2, y_2)\) are coordinates of two ends of diameter of circle. Note – If …
The normal at a point is the straight line which is perpendicular to the tangent to circle at the point of contact. Normal at a point of the circle passes through the center of circle. Here, you will learn how to find equation of normal to a circle with example. Equation of Normal to a …
When a straight line meet a circle on two coincident points then it is called the tangent to a circle. Here, you will learn condition of line to be a tangent to a circle and equation of tangent to a circle with example. Condition of Tangency : The line L = 0 touches the circle …
Equation of Tangent to a Circle – Condition of Tangency Read More »
Here, you will learn what is odd even functions and how to determine if given function is odd or even with example. Let’s begin – Odd Even Function : Let a function f(x) such that both x and -x are in its domain then If f(-x) = f(x) then f is said to be an …
The equation of normal to parabola in point form, slope form and parametric form are given below with examples. Equation of Normal to Parabola \(y^2 = 4ax\) (a) Point form : The equation of normal to the given parabola at its point (\(x_1, y_1\)) is y – \(y_1\) = \(-y_1\over 2a\)(x – \(x_1\)) Example : …
The equation of tangent to parabola in point form, slope form and parametric form are given below with examples. Condition of Tangency for Parabola : (a) The line y = mx + c meets the parabola \(y^2\) = 4ax in two points real, coincident or imaginary according as a >=< cm \(\implies\) condition of tangency …
Equation of Tangent to Hyperbola \(x^2\over a^2\) – \(y^2\over b^2\) = 1 (a) Point form : The equation of tangent to the given hyperbola at its point (\(x_1, y_1\)) is \(x{x_1}\over a^2\) – \(y{y_1}\over b^2\) = 1 Example : Find the equation of tangent to the hyperbola \(16x^2\) – \(9y^2\) = 144 at (5, 16/3). …
Equation of Normal to hyperbola : \(x^2\over a^2\) – \(y^2\over b^2\) = 1 (a) Point form : The equation of normal to the given hyperbola at the point P(\(x_1, y_1\)) is \(a^2x\over x_1\) + \(b^2y\over y_1\) = \(a^2+b^2\) = \(a^2e^2\) Example : Find the equation of normal to the hyperbola \(x^2\over 25\) – \(y^2\over 16\) = …
Equation of Tangent to Ellipse \(x^2\over a^2\) + \(y^2\over b^2\) = 1 : (a) Point form : The equation of tangent to the given ellipse at its point (\(x_1, y_1\)) is \(x{x_1}\over a^2\) + \(y{y_1}\over b^2\) = 1. Note – For general ellipse replace \(x^2\) by \(xx_1\), \(y^2\) by \(yy_1\), 2x by \(x + x_1\), …
Equation of Normal to ellipse : \(x^2\over a^2\) + \(y^2\over b^2\) = 1 (a) Point form : The Equation of normal to the given ellipse at (\(x_1, y_1\)) is \(a^2x\over x_1\) + \(b^2y\over y_1\) = \(a^2-b^2\) = \(a^2e^2\) Example : Find the normal to the ellipse \(9x^2+16y^2\) = 288 at the point (4,3). Solution : …
