Solution : Let the equation of circle be \((x-3)^2 + (y-0)^2 + \lambda y\) = 0 As it passes through (1, -2) \(\therefore\) \((1-3)^2 + (-2)^2 + \lambda (-2)\) = 0 \(\implies\) 4 + 4 โ 2\(\lambda\) = 0 \(\implies\) \(\lambda\) = 4 \(\therefore\) Equation of circle is \((x-3)^2 + y^2 + 4y\) = 0 โฆ
Solution : for parabola, \(y^2\) = 4x Let y = mx + \(1\over m\) is tangent line and it touches the parabola \(x^2\) = -32. \(\therefore\) \(x^2\) = -32(mx + \(1\over m\)) \(\implies\) \(x^2 + 32mx + {32\over m}\) = 0 Now, D = 0 because it touches the curve. \(\therefore\) \((32m)^2 โ 4.{32\over m}\) โฆ
The slope of the line touching both the parabolas \(y^2\) = 4x and \(x^2\) = -32 is Read More ยป
Solution : Let coordinates of the center of T be (0, k). Distance between their center is k + 1 = \(\sqrt{1 + (k โ 1)^2}\) where k is radius of circle T and 1 is radius of circle C,ย so sum of these is distance between their centers. \(\implies\) k + 1 =ย \(\sqrt{k^2 + โฆ
Solution : Since, (\(\alpha\), -\(\alpha\)) lie on 4ax+2ay+c=0 and 5bx+2by+d=0. \(\therefore\) 4a\(\alpha\) + 2a\(\alpha\) + c = 0 \(\implies\) \(\alpha\) = \(-c\over 2a\)ย โฆ..(i) Also, 5b\(\alpha\) โ 2b\(\alpha\) + d = 0 \(\implies\) \(\alpha\) = \(-d\over 3b\)ย ย โฆ..(i) from equation (i) and (ii), \(-c\over 2a\) = \(-d\over 3b\) 3bc = 2ad Similar Questions Find โฆ
Solution : Since PS is the median, so S is the mid point of triangle PQR. So, Coordinates of S = (\({7+6\over 2}, {3 โ 1\over 2}\)) = (\(13\over 2\), 1) Slope of line PS = (1 โ 2)/(13/2 โ 2) = \(-2\over 9\) Required equation passes through (1, -1) is y + 1 = โฆ
Solution : We have [\(\vec{a}\) + \(\vec{b}\) \(\vec{b}\) + \(\vec{c}\) \(\vec{c}\) + \(\vec{a}\)] = {(\(\vec{a}\) + \(\vec{b}\))\(\times\)(\(\vec{b}\) + \(\vec{c}\))}.(\(\vec{c}\) + \(\vec{a}\)) = {\(\vec{a}\)\(\times\)\(\vec{b}\) + \(\vec{a}\)\(\times\)\(\vec{c}\) + \(\vec{b}\)\(\times\)\(\vec{b}\) + \(\vec{b}\)\(\times\)\(\vec{c}\)}.(\(\vec{c}\) + \(\vec{a}\))ย {\(\vec{b}\)\(\times\)\(\vec{b}\) = 0} = {\(\vec{a}\)\(\times\)\(\vec{b}\) + \(\vec{a}\)\(\times\)\(\vec{c}\) + \(\vec{b}\)\(\times\)\(\vec{c}\)}.(\(\vec{c}\) + \(\vec{a}\)) = (\(\vec{a}\times\vec{b}\)).\(\vec{c}\) + (\(\vec{a}\times\vec{c}\)).\(\vec{c}\) + (\(\vec{b}\times\vec{c}\)).\(\vec{c}\) + (\(\vec{a}\times\vec{b}\)).\(\vec{a}\) + (\(\vec{a}\times\vec{c}\)).\(\vec{a}\) + (\(\vec{b}\times\vec{c}\)).\(\vec{a}\) = โฆ
Solution : \(\vec{a}\times\vec{b}\) = \(\vec{c}\) and \(\vec{b}\times\vec{c}\) = \(\vec{a}\) \(\implies\)ย \(\vec{c}\perp\vec{a}\) , \(\vec{c}\perp\vec{b}\) and \(\vec{a}\perp\vec{b}\), \(\vec{a}\perp\vec{c}\) \(\implies\)ย \(\vec{a}\perp\vec{b}\), \(\vec{b}\perp\vec{c}\) and \(\vec{c}\perp\vec{a}\) \(\implies\)ย \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are mutually perpendicular vectors. Again, \(\vec{a}\times\vec{b}\) = \(\vec{c}\) and \(\vec{b}\times\vec{c}\) = \(\vec{a}\) \(\implies\) |\(\vec{a}\times\vec{b}\)| = |\(\vec{c}\)| and |\(\vec{b}\times\vec{c}\)| = |\(\vec{a}\)| \(\implies\)ย \(|\vec{a}||\vec{b}|sin{\pi\over 2}\) = |\(\vec{c}\)| and \(|\vec{b}||\vec{c}|sin{\pi\over 2}\) = |\(\vec{a}\)|ย โฆ
Solution : Unit vectors perpendicular to \(\vec{a}\) & \(\vec{b}\) = \(\pm\)\(\vec{a}\times\vec{b}\over |\vec{a}\times\vec{b}|\) \(\therefore\)ย \(\vec{a}\times\vec{b}\) = \(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 1 & 2 & -2 \\ \end{vmatrix}\) = \(-5\hat{i} โ 5\hat{j} โ 5\hat{k}\) \(\therefore\) Unit Vectors = \(\pm\) \(-5\hat{i} โ 5\hat{j} โ 5\hat{k}\over 5\sqrt{3}\) Hence the required โฆ
Solution : cos2A + cos2B + cos2C = 2cos(A+B)cos(A-B) + cos2C = 2cos(\(3\pi\over 2\) โ C)cos(A-B) + cos2Cย \(\because\)ย A + B + C = \(3\pi\over 2\) = -2sinC cos(A-B) + 1 โ 2\(sin^2C\) = 1 โ 2sinC[cos(A-B)+sinC] = 1 โ 2sinC[cos(A-B) + sin(\(3\pi\over 2\)-(A+B))] = 1 โ 2sinC[cos(A-B)-cos(A+B)] = 1 โ 4sinA sinB sinC โฆ
If A + B + C = \(3\pi\over 2\), then cos2A + cos2B + cos2C is equal to Read More ยป
Solution : L.H.S. = \(2sin2xcos3x + sin2x\over{2cos3x.cos2x + 2cos3x + 2cos^2x}\) = \(sin2x[2cos3x+1]\over {2[cos3x(cos2x+1)+(cos^2x)]}\) = \(sin2x[2cos3x+1]\over {2[cos3x(2cos^2x)+(cos^2x)]}\) = \(sin2x[2cos3x+1]\over {2cos^2x(2cos3x+1)}\) = tanx Similar Questions Evaluate sin78 โ sin66 โ sin42 + sin6. If A + B + C = \(3\pi\over 2\), then cos2A + cos2B + cos2C is equal to Find the maximum value of โฆ
\(sin5x + sin2x โ sinx\over {cos5x + 2cos3x + 2cos^x + cosx}\) is equal to Read More ยป
