Solution : The locus of the intersection of tangents which are at right angles is known as director circle of the hyperbola. The equation to the director circle is : \(x^2+y^2\) = \(a^2-b^2\) If \(b^2\) < \(a^2\), this circle is real ; If \(b^2\) = \(a^2\) the radius of the circle is zero & it โฆ
What is the Equation of Director Circle of Hyperbola ? Read More ยป
Solution : The equation x = acos\(\theta\) & y = bsin\(\theta\) together represent the parametric equation of ellipse \({x_1}^2\over a^2\) + \({y_1}^2\over b^2\) = 1, where \(\theta\) is a parameter. Note that if P(\(\theta\)) = (acos\(\theta\), bsin\(\theta\)) is on the ellipse then ; Q(\(\theta\)) = (acos\(\theta\), bsin\(\theta\)) is on auxilliary circle. A circle described on โฆ
Solution : Let y = \(x^{sinx}\). Then, Taking log both sides, log y = sin x.log x \(\implies\) y = \(e^{sin x.log x}\) By using logarithmic differentiation, On differentiating both sides with respect to x, we get \(dy\over dx\) = \(e^{sin x.log x}\)\(d\over dx\)(sin x.log x) \(\implies\) \(dy\over dx\) = \(x^{sin x}{log x {d\over dx}(sin โฆ
Solution : Since by deleting a single term from an infinite series, it remains same. Therefore, the given function may be written as y = \(\sqrt{sin x + y}\) Squaring on both sides, \(\implies\)ย \(y^2\)ย = sin x + y By using differentiation of infinite series, Differentiating both sides with respect to x, 2y \(dy\over โฆ
Solution : We have, x = a{cos t + \({1\over 2} log tan^2 {t\over 2}\)} and y = a sin t \(\implies\) x = a{cos t + \({1\over 2} \times 2 log tan{t\over 2}\)} and y = a sin t \(\implies\) x = a{cos t + {\(log tan{t\over 2}\)} and y = a sin t โฆ
Solution : The integration of \(e^x\) with respect to x is \(e^x\) + C. Since \(d\over dx\) \(e^x\) = \(e^x\) dx On integrating both sides, we get \(\int\) \(e^x\) dx = \(e^x\) Hence, the integration of \(e^x\) is \(e^x\) + C
Solution : | A | = \(\begin{vmatrix} 3 & -2 & 4 \\ 1 & 2 & 1 \\ 0 & 1 & -1 \end{vmatrix}\) By using 3ร3 determinant formula, \(\implies\) | A | = \(3\begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix}\) โ \((-2)\begin{vmatrix} 1 & 1 \\ 0 & -1 \end{vmatrix}\) + โฆ
Solution : Let | A | = \(\begin{vmatrix} sinx & cosx \\ -cosx & sinx \end{vmatrix}\) By using 2ร2 determinant formula, we obtain | A | = ( \(sin^2x\)) โ (\(-cos^2x\)) = \(sin^2x\) + \(cos^2x\) = 1
Solution : Let \(C_1\) be the center of circle \(x^2 + y^2\) = 1 i.e.ย \(C_1\) = (0, 0) And \(C_2\) be the center of circle \(x^2 + y^2 โ 2x โ 6y + 6\) = 0 i.e. \(C_2\) = (1, 3) Let \(r_1\) be the radius of first circle and \(r_2\) be the radius โฆ
Solution : The given expression is \((3x โ {x^3\over 6})^7\). Here n = 7, which is an odd number. By using middle terms in binomial expansion formula, So, \(({7 + 1\over 2}\)) th and \(({7 + 1\over 2} + 1)\) th i.e.ย 4th and 5th terms are two middle terms. Now, \(T_{4}\) = \(T_{3 + โฆ
Find the middle term in the expansion of \((3x โ {x^3\over 6})^7\). Read More ยป
