Parabola Questions

Solution : The given parabola is of form \(y^2\) = 4ax. On comparing, we have 4a = 12 i.e a = 3. We know that the focal distance of any point (x, y) on \(y^2\) = 4ax is x + a. Let the given point on the parabola \(y^2\) = 12 x be (x, y). […]

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 +

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. \(\therefore\) \((32m)^2 – 4.{32\over m}\)

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

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 : The equation of tangent in slope form to \(y^2\) = 4ax is y = mx + \(a\over m\) Now, if it is common to both parabola, it also lies on second parabola then \(x^2\) = 4a(mx + \(a\over m\)) \(mx^2 – 4am^2 – 4a^2\) = 0 has equal roots. then its discriminant is

Solution : The Equation of tangent to the parabola having slope ‘m’, is y = mx + \(a\over m\) ,  (m \(\ne\) 0) and point of contact  is (\(a\over m^2\), \(2a\over m\)). Similar Questions The sum of the slopes of the tangent of the parabola \(y^2\)=4ax drawn from the point (2,3) is Find the locus