If p is the length of the perpendicular from the origin to the line \(x\over a\) + \(y\over b\) = 1, then prove that \(1\over p^2\) = \(1\over a^2\) + \(1\over b^2\)

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Solution : The given line is bx + ay – ab = 0 ………….(i) It is given that p = Length of the perpendicular from the origin to line (i) \(\implies\) p = \(|b(0) + a(0) – ab|\over {\sqrt{b^2+a^2}}\) = \(ab\over \sqrt{a^2+b^2}\) \(\implies\) \(p^2\) = \(a^2b^2\over a^2+b^2\) \(\implies\) \(1\over p^2\) = \(a^2+b^2\over a^2b^2\) \(\implies\) \(1\over […]

If p is the length of the perpendicular from the origin to the line \(x\over a\) + \(y\over b\) = 1, then prove that \(1\over p^2\) = \(1\over a^2\) + \(1\over b^2\) Read More »

Find the distance between the line 12x – 5y + 9 = 0 and the point (2,1)

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Solution : We have line 12x – 5y + 9 = 0 and the point (2,1) Required distance = |\(12*2 – 5*1 + 9\over {\sqrt{12^2 + (-5)^2}}\)| = \(|24-5+9|\over 13\) = \(28\over 13\) Similar Questions If p is the length of the perpendicular from the origin to the line \(x\over a\) + \(y\over b\) =

Find the distance between the line 12x – 5y + 9 = 0 and the point (2,1) Read More »

Let \(a_n\) be the nth term of an AP. If \(\sum_{r=1}^{100}\) \(a_{2r}\) = \(\alpha\) and \(\sum_{r=1}^{100}\) \(a_{2r-1}\) = \(\beta\), then the common difference of the AP is

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Solution : Given, \(a_2 + a_4 + a_6 + …… + a_{200}\) = \(\alpha\)      ………(i) and \(a_1 + a_3 + a_5 + ….. + a_{199}\) = \(\beta\)           ………(ii) On subtracting equation (ii) from equation (i), we get (\(a_2 – a_1\)) + (\(a_4 – a_3\)) + ……… + (\(a_{200} –

Let \(a_n\) be the nth term of an AP. If \(\sum_{r=1}^{100}\) \(a_{2r}\) = \(\alpha\) and \(\sum_{r=1}^{100}\) \(a_{2r-1}\) = \(\beta\), then the common difference of the AP is Read More »

A man saves Rs 200 in each of the first three months of his service. In each of the subsequent months, his saving increases by Rs 40 more than the saving of immediately previous month. His total saving from the start of service will be Rs 11040 after

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Solution : Let the time taken to save Rs 11040 be (n + 3) months. for first three months, he saves Rs 200 each month. In (n + 3) months, 3 \(\times\) 200 + \(n\over 2\) { 2(240) + (n – 1) \(\times\) 40 } = 11040 \(\implies\)  600 + \(n\over 2\) {40(12+ n –

A man saves Rs 200 in each of the first three months of his service. In each of the subsequent months, his saving increases by Rs 40 more than the saving of immediately previous month. His total saving from the start of service will be Rs 11040 after Read More »

If 100 times the 100th term of an AP with non-zero common difference equal to the 50 times its 50th term, then the 150th term of AP is

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Solution : Let a be the first term and d (d \(\ne\) 0) be the common difference of the given AP, then \(T_{100}\) = a + (100 – 1)d = a + 99d \(T_{50}\) = a + (50 – 1)d = a + 49d \(T_{150}\) = a + (150 – 1)d = a + 149d

If 100 times the 100th term of an AP with non-zero common difference equal to the 50 times its 50th term, then the 150th term of AP is Read More »

Angle between asymptotes of hyperbola xy=8 is

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Solution : Since given hyperbola xy = 8 is rectangular hyperbola. And eccentricity of rectangular hyperbola is \(\sqrt{2}\) Angle between asymptotes of hyperbola is \(2sec^{-1}(e)\) \(\implies\) \(\theta\) = \(2sec^{-1}(\sqrt{2})\) \(\implies\) \(\theta\) = \(2sec^{-1}(sec 45)\) \(\implies\) \(\theta\) = 2(45) = 90 Similar Questions Find the normal to the hyperbola \(x^2\over 16\) – \(y^2\over 9\) = 1

Angle between asymptotes of hyperbola xy=8 is Read More »

The sum of the slopes of the tangent of the parabola \(y^2\)=4ax drawn from the point (2,3) is

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Solution : The equation of tangent to the parabola \(y^2\) = 4ax is y = mx + \(a\over m\). Since it is drawn from point (2,3) Therefore it lies on tangent y = mx + \(a\over m\). \(\implies\) 3 = 2m + \(a\over m\) \(\implies\) 3m = 2\(m^2\) + a \(\implies\)  2\(m^2\) – 3m +

The sum of the slopes of the tangent of the parabola \(y^2\)=4ax drawn from the point (2,3) is Read More »

If x, y and z are in AP and \(tan^{-1}x\), \(tan^{-1}y\) and \(tan^{-1}z\) are also in AP, then

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Solution : Since, x, y and z are in AP \(\therefore\)   2y = x + z Also,  \(tan^{-1}x\), \(tan^{-1}y\) and \(tan^{-1}z\) are in AP \(\therefore\)   2\(tan^{-1}y\) =  \(tan^{-1}x\) +  \(tan^{-1}z\) \(\implies\) \(tan^{-1}({2y\over {1 – y^2}})\) = \(tan^{-1}({x + z\over {1 – xz}})\) \(\implies\) \(x + z\over {1 – y^2}\) = \(x + z\over {1 –

If x, y and z are in AP and \(tan^{-1}x\), \(tan^{-1}y\) and \(tan^{-1}z\) are also in AP, then Read More »

The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, ……. , is

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Solution : 0.7 + 0.77 + 0.777 + …… + upto 20 terms = \(7\over 10\) + \(77\over 10^2\) + \(777\over 10^3\) +  ….. + upto 20 terms = 7[ \(1\over 10\) +  \(11\over 10^2\) + \(111\over 10^3\) +  ….. + upto 20 terms ] = \(7\over 9\)[ \(9\over 10\) +  \(99\over 100\) + \(999\over

The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, ……. , is Read More »