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 : […]
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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
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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).
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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 \(x^2\over 25\) – \(y^2\over 16\) =
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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 \(x + x_1\),
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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 :
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Let \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) be three vectors. Then the scalar \((\vec{a}\times \vec{b}).\vec{c}\) is called the scalar triple product of \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) and is denoted by [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)]. Thus, we have [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] = \((\vec{a}\times \vec{b}).\vec{c}\) For three vectors \(\vec{a}\), \(\vec{b}\) & \(\vec{c}\), it is also defined as : (\(\vec{a}\times\vec{b}\)).\(\vec{c}\) = \(|\vec{a}||\vec{b}||\vec{c}|sin\theta
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Here you will learn what is domain, codomain and range of function and how to find domain and range of function with example. Let’s begin – Let f : A \(\rightarrow\) B is a function. Then, the set A is known as the domain of f and the set B is known as the codomain
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Formula for Integration by Parts If u and v are two functions of x, then the formula for integration by parts is – \(\int\) u.v dx = u \(\int\) v dx – \(\int\)[\(du\over dx\).\(\int\)v dx]dx i.e The integral of the product of two functions = (first function) \(\times\) (Integral of Second function) – Integral of
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Earlier we find Area of Triangle by using the formula – Area of Triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\) where s = \(a+b+c\over 2\) and a, b, c are the sides of the triangle. we have used this heron’s formula to find the area of a triangle when the lengths of its sides are given. Here, we will
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