Solution : Let P(h,k) be the mid point of chord of the parabola \(y^2\) = 4ax, so equation of chord is yk – 2a(x+h) = \(k^2\) – 4ah. Since it passes through (p,q) \(\therefore\) qk – 2a(p+h) = \(k^2\) – 4ah \(\therefore\) Required locus is \(y^2\) – 2ax – qy + 2ap = 0 Similar […]
Solution : Equation of tangent to the parabola \(y^2\) = 9x is y = mx + \(9\over 4m\) Since it passes through (4,10) \(\therefore\) 10 = 4m + \(9\over 4m\) \(\implies\) 16\(m^2\) – 40m + 9 = 0 m = \(1\over 4\), \(9\over 4\) \(\therefore\) Equation of tangent’s are y = \(x\over 4\) + 9
Solution : \(\because\) Point (k-1, k) lies inside the parabola \(y^2\) = 4x. \(\therefore\) \({y_1}^2 – 4ax_1\) < 0 \(\implies\) \(k^2\) – 4(k-1) < 0 \(\implies\) \(k^2\) – 4k + 4 < 0 \((k-2)^2\) < 0 \(\implies\) k \(\in\) \(\phi\) Similar Questions The slope of the line touching both the parabolas \(y^2\) = 4x and
Find the value of k for which the point (k-1, k) lies inside the parabola \(y^2\) = 4x. Read More »
Solution : The length of latus rectum = 2 x perpendicular from focus to the directrix = 2 x |\({2-4(3)+3}\over {\sqrt{1+16}}\)| = \(14\over \sqrt{17}\) Similar Questions The slope of the line touching both the parabolas \(y^2\) = 4x and \(x^2\) = -32 is Find the locus of middle point of the chord of the parabola
Solution : We have \(2^{x+2}\) > \(2^{-2/x}\) Since the base 2 > 1, we have x + 2 > \(-2\over x\) (the sign of inequality is retained) Now, x + 2 + \(-2\over x\) > 0 \(\implies\) \({x^2 + 2x + 2}\over x\) > 0 \(\implies\) \(({x+1})^2 + 1\over x\) > 0 \(\implies\) x \(\in\)
Solve for x : \(2^{x+2}\) > \(({1\over 4})^{1/x}\) Read More »
Solution : \(log_a x\) = p \(\implies\) \(a^p\) = x \(\implies\) a = \(x^{1/p}\) Similarly \(b^q\) = \(x^2\) \(\implies\) b = \(x^{2/q}\) Now, \(log_x \sqrt{ab}\) = \(log_x \sqrt{x^{1/p}x^{2/q}}\) = \(log_x x^{({1\over p}+{2\over q}){1\over 2}}\) = \(1\over {2p}\) + \(1\over q\). Similar Questions Solve for x : \(2^{x + 2}\) > \(({1\over 4})^{1\over x}\). Evaluate the
If \(log_a x\) = p and \(log_b {x^2}\) = q then \(log_x \sqrt{ab}\) is equal to Read More »
Solution : \(log_e x\) – \(log_e y\) = a \(\implies\) \(log_e {x\over y}\) = a \(\implies\) \(x\over y\) = \(e^a\) \(log_e y\) – \(log_e z\) = b \(\implies\) \(log_e {y\over z}\) = b \(\implies\) \(y\over z\) = \(e^b\) \(log_e z\) – \(log_e x\) = c \(\implies\) \(log_e {z\over x}\) = c \(\implies\) \(z\over x\) =
Solution : Here f(x) = \({7x^2+1\over 5x^2-1}\) \(\phi\)(x) = \({x^5\over {1-x^3}}\) = \(x^2x^3\over 1-x^3\) = \(x^2\over {1\over x^3}-1\) \(\therefore\) \(\displaystyle{\lim_{x \to \infty}}\) f(x) = \(7\over 5\) & \(\displaystyle{\lim_{x \to \infty}}\) \(\phi\)(x) \(\rightarrow\) – \(\infty\) \(\implies\) \(\displaystyle{\lim_{x \to \infty}}\) \((f(x))^{\phi (x)}\) = \(({7\over 5})^{-\infty}\) = 0 Similar Questions Evaluate : \(\displaystyle{\lim_{x \to 0}}\) \(x^3 cotx\over {1-cosx}\)
Solution : \(\displaystyle{\lim_{x \to 0}}\) \(xln(1+2tanx)\over 1-cosx\) = \(\displaystyle{\lim_{x \to 0}}\) \(xln(1+2tanx)\over {1-cosx\over x^2}.x^2\).\(2tanx\over 2tanx\) = 4 Similar Questions Evaluate : \(\displaystyle{\lim_{x \to 0}}\) \(x^3 cotx\over {1-cosx}\) Evaluate : \(\displaystyle{\lim_{x \to \infty}}\) \(({7x^2+1\over 5x^2-1})^{x^5\over {1-x^3}}\) Evaluate the limit : \(\displaystyle{\lim_{x \to \infty}}\) \(x^2 + x + 1\over {3x^2 + 2x – 5}\) Evaluate : \(\displaystyle{\lim_{x
Evaluate : \(\displaystyle{\lim_{x \to 0}}\) \(xln(1+2tanx)\over 1-cosx\) Read More »
Solution : \(\displaystyle{\lim_{x \to 0}}\) \(2(sin(2+x)-sin2)+xsin(2+x)\over x\) = \(\displaystyle{\lim_{x \to 0}}\)(\(2.2.cos(2+{x\over 2})sin{x\over 2}\over x\) + sin(2+x)) = \(\displaystyle{\lim_{x \to 0}}\)\(2cos(2+{x\over 2})sin{x\over 2}\over {x\over 2}\) + \(\displaystyle{\lim_{x \to 0}}\)sin(2+x) = 2cos2 + sin2 Similar Questions Evaluate : \(\displaystyle{\lim_{x \to 0}}\) \(x^3 cotx\over {1-cosx}\) Evaluate : \(\displaystyle{\lim_{x \to \infty}}\) \(({7x^2+1\over 5x^2-1})^{x^5\over {1-x^3}}\) Evaluate : \(\displaystyle{\lim_{x \to 0}}\)
Evaluate : \(\displaystyle{\lim_{x \to 0}}\) \((2+x)sin(2+x)-2sin2\over x\) Read More »
